1. A solid is a (figure) having length and breadth and depth.
2. The extremity of a solid (is) a surface.
3. A straight-line is at right-angles to a plane when it makes right-angles with all of the straight-lines joined to it which are also in the plane.
4. A plane is at right-angles to a(nother) plane when (all of) the straight-lines drawn in one of the planes, at right-angles to the common section of the planes, are at right-angles to the remaining plane.
5. The inclination of a straight-line to a plane is the angle contained by the drawn and standing (straight-lines), when a perpendicular is lead to the plane from the end of the (standing) straight-line raised (out of the plane), and a straight-line is (then) joined from the point (so) generated to the end of the (standing) straight-line (lying) in the plane.
6. The inclination of a plane to a(nother) plane is the acute angle contained by the (straight-lines), (one) in each of the planes, drawn at right-angles to the common segment (of the planes), at the same point.
7. A plane is said to have been similarly inclined to a plane, as another to another, when the aforementioned angles of inclination are equal to one another.
8. Parallel planes are those which do not meet (one another).
9. Similar solid figures are those contained by equal numbers of similar planes (which are similarly arranged).
10. But equal and similar solid figures are those contained by similar planes equal in number and in magnitude (which are similarly arranged).
11. A solid angle is the inclination (constituted) by more than two lines joining one another (at the same point), and not being in the same surface, to all of the lines.
Otherwise, a solid angle is that contained by more than two plane angles, not being in the same plane, and constructed at one point.
12. A pyramid is a solid figure, contained by planes, (which is) constructed from one plane to one point.
13. A prism is a solid figure, contained by planes, of which the two opposite (planes) are equal, similar, and parallel, and the remaining (planes are) parallelograms.
14. A sphere is the figure enclosed when, the diameter of a semicircle remaining (fixed), the semicircle is carried around, and again established at the same (position) from which it began to be moved.
15. And the axis of the sphere is the fixed straight-line about which the semicircle is turned.
16. And the center of the sphere is the same as that of the semicircle.
17. And the diameter of the sphere is any straight-line which is drawn through the center and terminated in both directions by the surface of the sphere.
18. A cone is the figure enclosed when, one of the sides of a right-angled triangle about the right-angle remaining (fixed), the triangle is carried around, and again established at the same (position) from which it began to be moved.
And if the fixed straight-line is equal to the remaining (straight-line) about the right-angle, (which is) carried around, then the cone will be right-angled, and if less, obtuse-angled, and if greater, acute-angled.
19. And the axis of the cone is the fixed straight-line about which the triangle is turned.
20. And the base (of the cone is) the circle described by the (remaining) straight-line (about the right-angle which is) carried around (the axis).
21. A cylinder is the figure enclosed when, one of the sides of a right-angled parallelogram about the right-angle remaining (fixed), the parallelogram is carried around, and again established at the same (position) from which it began to be moved.
22. And the axis of the cylinder is the stationary straight-line about which the parallelogram is turned.
23. And the bases (of the cylinder are) the circles described by the two opposite sides (which are) carried around.
24. Similar cones and cylinders are those for which the axes and the diameters of the bases are proportional.
25. A cube is a solid figure contained by six equal squares.
26. An octahedron is a solid figure contained by eight equal and equilateral triangles.
27. An icosahedron is a solid figure contained by twenty equal and equilateral triangles.
28. A dodecahedron is a solid figure contained by twelve equal, equilateral, and equiangular pentagons.
Proposition 1
Some part of a straight-line cannot be in a reference plane, and some part in a more elevated (plane).
For, if possible, let some part, AB, of the straight-line ABC be in a reference plane, and some part, BC, in a more elevated (plane).
In the reference plane, there will be some straight-line continuous with, and straight-on to, AB.
Let it be BD.
Thus, AB is a common segment of the two (different) straight-lines ABC and ABD.
The very thing is impossible, inasmuch as if we draw a circle with center B and radius AB then the diameters ( ABD and ABC) will cut off unequal circumferences of the circle.
Thus, some part of a straight-line cannot be in a reference plane, and (some part) in a more elevated (plane).
(Which is) the very thing it was required to show.
Proposition 2
If two straight-lines cut one another then they are in one plane, and every triangle (formed using segments of both lines) is in one plane.
For let the two straight-lines AB and CD have cut one another at point E.
I say that AB and CD are in one plane, and that every triangle (formed using segments of both lines) is in one plane.
For let the random points F and G have been taken on EC and EB (respectively).
And let CB and FG have been joined, and let FH and GK have been drawn across.
I say, first of all, that triangle ECB is in one (reference) plane.
For if part of triangle ECB, either FHC or GBK, is in the reference [plane], and the remainder in a different (plane) then a part of one the straight-lines EC and EB will also be in the reference plane, and (a part) in a different (plane).
And if the part FCBG of triangle ECB is in the reference plane, and the remainder in a different (plane) then parts of both of the straight-lines EC and EB will also be in the reference plane, and (parts) in a different (plane).
The very thing was shown to be absurb [Prop. 11.1].
Thus, triangle ECB is in one plane.
And in whichever (plane) triangle ECB is (found), in that (plane) EC and EB (will) each also (be found).
And in whichever (plane) EC and EB (are) each (found), in that (plane) AB and CD (will) also (be found) [Prop. 11.1].
Thus, the straight-lines AB and CD are in one plane, and every triangle (formed using segments of both lines) is in one plane.
(Which is) the very thing it was required to show.
Proposition 3
If two planes cut one another then their common section is a straight-line.
For let the two planes AB and BC cut one another, and let their common section be the line DB.
I say that the line DB is straight.
For, if not, let the straight-line DEB have been joined from D to B in the plane AB, and the straight-line DFB in the plane BC.
So two straight-lines, DEB and DFB, will have the same ends, and they will clearly enclose an area.
The very thing (is) absurd.
Thus, DEB and DFB are not straight-lines.
So, similarly, we can show than no other straight-line can be joined from D to B except DB, the common section of the planes AB and BC.
Thus, if two planes cut one another then their common section is a straight-line.
(Which is) the very thing it was required to show.
Proposition 4
If a straight-line is set up at right-angles to two straight-lines cutting one another, at the common point of section, then it will also be at right-angles to the plane (passing) through them (both).
For let some straight-line EF have (been) set up at right-angles to two straight-lines, AB and CD, cutting one another at point E, at E.
I say that EF is also at right-angles to the plane (passing) through AB and CD.
For let AE, EB, CE and ED have been cut off from (the two straight-lines so as to be) equal to one another.
And let GEH have been drawn, at random, through E (in the plane passing through AB and CD).
And let AD and CB have been joined.
And, furthermore, let FA, FG, FD, FC, FH, and FB have been joined from the random (point) F (on EF).
For since the two (straight-lines) AE and ED are equal to the two (straight-lines) CE and EB, and they enclose equal angles [Prop. 1.15], the base AD is thus equal to the base CB, and triangle AED will be equal to triangle CEB [Prop. 1.4].
Hence, the angle DAE [is] equal to the angle EBC.
And the angle AEG (is) also equal to the angle BEH [Prop. 1.15].
So AGE and BEH are two triangles having two angles equal to two angles, respectively, and one side equal to one side---(namely), those by the equal angles, AE and EB.
Thus, they will also have the remaining sides equal to the remaining sides [Prop. 1.26].
Thus, GE (is) equal to EH, and AG to BH
And since AE is equal to EB, and FE is common and at right-angles, the base FA is thus equal to the base FB [Prop. 1.4].
So, for the same (reasons), FC is also equal to FD.
And since AD is equal to CB, and FA is also equal to FB, the two (straight-lines) FA and AD are equal to the two (straight-lines) FB and BC, respectively.
And the base FD was shown (to be) equal to the base FC.
Thus, the angle FAD is also equal to the angle FBC [Prop. 1.8].
And, again, since AG was shown (to be) equal to BH, but FA (is) also equal to FB, the two (straight-lines) FA and AG are equal to the two (straight-lines) FB and BH (respectively).
And the angle FAG was shown (to be) equal to the angle FBH.
Thus, the base FG is equal to the base FH [Prop. 1.4].
And, again, since GE was shown (to be) equal to EH, and EF (is) common, the two (straight-lines) GE and EF are equal to the two (straight-lines) HE and EF (respectively).
And the base FG (is) equal to the base FH.
Thus, the angle GEF is equal to the angle HEF [Prop. 1.8].
Each of the angles GEF and HEF (are) thus right-angles [Def. 1.10].
Thus, FE is at right-angles to GH, which was drawn at random through E (in the reference plane passing though AB and CD).
So, similarly, we can show that FE will make right-angles with all straight-lines joined to it which are in the reference plane.
And a straight-line is at right-angles to a plane when it makes right-angles with all straight-lines joined to it which are in the plane [Def. 11.3].
Thus, FE is at right-angles to the reference plane.
And the reference plane is that (passing) through the straight-lines AB and CD.
Thus, FE is at right-angles to the plane (passing) through AB and CD.
Thus, if a straight-line is set up at right-angles to two straight-lines cutting one another, at the common point of section, then it will also be at right-angles to the plane (passing) through them (both).
(Which is) the very thing it was required to show.
Proposition 5
If a straight-line is set up at right-angles to three straight-lines cutting one another, at the common point of section, then the three straight-lines are in one plane.
For let some straight-line AB have been set up at right-angles to three straight-lines BC, BD, and BE, at the (common) point of section B.
I say that BC, BD, and BE are in one plane.
For (if) not, and if possible, let BD and BE be in the reference plane, and BC in a more elevated (plane).
And let the plane through AB and BC have been produced.
So it will make a straight-line as a common section with the reference plane [Def. 11.3].
Let it make BF.
Thus, the three straight-lines AB, BC, and BF are in one plane—(namely), that drawn through AB and BC.
And since AB is at right-angles to each of BD and BE, AB is thus also at right-angles to the plane (passing) through BD and BE [Prop. 11.4].
And the plane (passing) through BD and BE is the reference plane.
Thus, AB is at right-angles to the reference plane.
Hence, AB will also make right-angles with all straight-lines joined to it which are also in the reference plane [Def. 11.3].
And BF, which is in the reference plane, is joined to it.
Thus, the angle ABF is a right-angle.
And ABC was also assumed to be a right-angle.
Thus, angle ABF (is) equal to ABC.
And they are in one plane.
The very thing is impossible.
Thus, BC is not in a more elevated plane.
Thus, the three straight-lines BC, BD, and BE are in one plane.
Thus, if a straight-line is set up at right-angles to three straight-lines cutting one another, at the (common) point of section, then the three straight-lines are in one plane.
(Which is) the very thing it was required to show.
Proposition 6
If two straight-lines are at right-angles to the same plane then the straight-lines will be parallel.
For let the two straight-lines AB and CD be at right-angles to a reference plane.
I say that AB is parallel to CD.
For let them meet the reference plane at points B and D (respectively). And let the straight-line BD have been joined.
And let DE have been drawn at right-angles to BD in the reference plane.
And let DE be made equal to AB.
And let BE, AE, and AD have been joined.
And since AB is at right-angles to the reference plane, it will [thus] also make right-angles with all straight-lines joined to it which are in the reference plane [Def. 11.3].
And BD and BE, which are in the reference plane, are each joined to AB.
Thus, each of the angles ABD and ABE are right-angles.
So, for the same (reasons), each of the angles CDB and CDE are also right-angles.
And since AB is equal to DE, and BD (is) common, the two (straight-lines) AB and BD are equal to the two (straight-lines) ED and DB (respectively).
And they contain right-angles.
Thus, the base AD is equal to the base BE [Prop. 1.4].
And since AB is equal to DE, and AD (is) also (equal) to BE, the two (straight-lines) AB and BE are thus equal to the two (straight-lines) ED and DA (respectively).
And their base AE (is) common.
Thus, angle ABE is equal to angle EDA [Prop. 1.8].
And ABE (is) a right-angle.
Thus, EDA (is) also a right-angle.
ED is thus at right-angles to DA.
And it is also at right-angles to each of BD and DC.
Thus, ED is standing at right-angles to the three straight-lines BD, DA, and DC at the (common) point of section.
Thus, the three straight-lines BD, DA, and DC are in one plane [Prop. 11.5].
And in which(ever) plane DB and DA (are found), in that (plane) AB (will) also (be found).
For every triangle is in one plane [Prop. 11.2].
And each of the angles ABD and BDC is a right-angle.
Thus, AB is parallel to CD [Prop. 1.28].
Thus, if two straight-lines are at right-angles to the same plane then the straight-lines will be parallel.
(Which is) the very thing it was required to show.
Proposition 7
If there are two parallel straight-lines, and random points are taken on each of them, then the straight-line joining the two points is in the same plane as the parallel (straight-lines).
Let AB and CD be two parallel straight-lines, and let the random points E and F have been taken on each of them (respectively).
I say that the straight-line joining points E and F is in the same (reference) plane as the parallel (straight-lines).
For (if) not, and if possible, let it be in a more elevated (plane), such as EGF.
And let a plane have been drawn through EGF.
So it will make a straight cutting in the reference plane [Prop. 11.3].
Let it make EF.
Thus, two straight-lines (with the same end-points), EGF and EF, will enclose an area.
The very thing is impossible.
Thus, the straight-line joining E to F is not in a more elevated plane.
The straight-line joining E to F is thus in the plane through the parallel (straight-lines) AB and CD.
Thus, if there are two parallel straight-lines, and random points are taken on each of them, then the straight-line joining the two points is in the same plane as the parallel (straight-lines).
(Which is) the very thing it was required to show.
Proposition 8
If two straight-lines are parallel, and one of them is at right-angles to some plane, then the remaining (one) will also be at right-angles to the same plane.
Let AB and CD be two parallel straight-lines, and let one of them, AB, be at right-angles to a reference plane.
I say that the remaining (one), CD, will also be at right-angles to the same plane.
For let AB and CD meet the reference plane at points B and D (respectively).
And let BD have been joined.
AB, CD, and BD are thus in one plane [Prop. 11.7].
Let DE have been drawn at right-angles to BD in the reference plane, and let DE be made equal to AB, and let BE, AE, and AD have been joined.
And since AB is at right-angles to the reference plane, AB is thus also at right-angles to all of the straight-lines joined to it which are in the reference plane [Def. 11.3].
Thus, the angles ABD and ABE [are] each right-angles.
And since the straight-line BD has met the parallel (straight-lines) AB and CD, the (sum of the) angles ABD and CDB is thus equal to two right-angles [Prop. 1.29].
And ABD (is) a right-angle.
Thus, CDB (is) also a right-angle.
CD is thus at right-angles to BD.
And since AB is equal to DE, and BD (is) common, the two (straight-lines) AB and BD are equal to the two (straight-lines) ED and DB (respectively).
And angle ABD (is) equal to angle EDB.
For each (is) a right-angle.
Thus, the base AD (is) equal to the base BE [Prop. 1.4].
And since AB is equal to DE, and BE to AD, the two (sides) AB, BE are equal to the two (sides) ED, DA, respectively.
And their base AE is common.
Thus, angle ABE is equal to angle EDA [Prop. 1.8].
And ABE (is) a right-angle.
EDA (is) thus also a right-angle.
Thus, ED is at right-angles to AD.
And it is also at right-angles to DB.
Thus, ED is also at right-angles to the plane through BD and DA [Prop. 11.4].
And ED will thus make right-angles with all of the straight-lines joined to it which are also in the plane through BDA.
And DC is in the plane through BDA, inasmuch as AB and BD are in the plane through BDA [Prop. 11.2], and in which(ever plane) AB and BD (are found), DC is also (found).
Thus, ED is at right-angles to DC.
Hence, CD is also at right-angles to DE.
And CD is also at right-angles to BD.
Thus, CD is standing at right-angles to two straight-lines, DE and DB, which meet one another, at the (point) of section, D.
Hence, CD is also at right-angles to the plane through DE and DB [Prop. 11.4].
And the plane through DE and DB is the reference (plane).
CD is thus at right-angles to the reference plane.
Thus, if two straight-lines are parallel, and one of them is at right-angles to some plane, then the remaining (one) will also be at right-angles to the same plane.
(Which is) the very thing it was required to show.
Proposition 9
(Straight-lines) parallel to the same straight-line, and which are not in the same plane as it, are also parallel to one another.
For let AB and CD each be parallel to EF, not being in the same plane as it.
I say that AB is parallel to CD.
For let some point G have been taken at random on EF.
And from it let GH have been drawn at right-angles to EF in the plane through EF and AB.
And let GK have been drawn, again at right-angles to EF, in the plane through FE and CD.
And since EF is at right-angles to each of GH and GK, EF is thus also at right-angles to the plane through GH and GK [Prop. 11.4].
And EF is parallel to AB.
Thus, AB is also at right-angles to the plane through HGK [Prop. 11.8].
So, for the same (reasons), CD is also at right-angles to the plane through HGK.
Thus, AB and CD are each at right-angles to the plane through HGK.
And if two straight-lines are at right–angles to the same plane then the straight-lines are parallel [Prop. 11.6].
Thus, AB is parallel to CD.
(Which is) the very thing it was required to show.
Proposition 10
If two straight-lines joined to one another are (respectively) parallel to two straight-lines joined to one another, (but are) not in the same plane, then they will contain equal angles.
For let the two straight-lines joined to one another, AB and BC, be (respectively) parallel to the two straight-lines joined to one another, DE and EF, (but) not in the same plane.
I say that angle ABC is equal to (angle) DEF.
For let BA, BC, ED, and EF have been cut off (so as to be, respectively) equal to one another.
And let AD, CF, BE, AC, and DF have been joined.
And since BA is equal and parallel to ED, AD is thus also equal and parallel to BE [Prop. 1.33].
So, for the same reasons, CF is also equal and parallel to BE.
Thus, AD and CF are each equal and parallel to BE.
And straight-lines parallel to the same straight-line, and which are not in the same plane as it, are also parallel to one another [Prop. 11.9].
Thus, AD is parallel and equal to CF.
And AC and DF join them.
Thus, AC is also equal and parallel to DF [Prop. 1.33].
And since the two (straight-lines) AB and BC are equal to the two (straight-lines) DE and EF (respectvely), and the base AC (is) equal to the base DF, the angle ABC is thus equal to the (angle) DEF [Prop. 1.8].
Thus, if two straight-lines joined to one another are (respectively) parallel to two straight-lines joined to one another, (but are) not in the same plane, then they will contain equal angles.
(Which is) the very thing it was required to show.
Proposition 11
To draw a perpendicular straight-line from a given raised point to a given plane.
Let A be the given raised point, and the given plane the reference (plane).
So, it is required to draw a perpendicular straight-line from point A to the reference plane.
Let some random straight-line BC have been drawn across in the reference plane, and let the (straight-line) AD have been drawn from point A perpendicular to BC [Prop. 1.12].
If, therefore, AD is also perpendicular to the reference plane then that which was prescribed will have occurred.
And, if not, let DE have been drawn in the reference plane from point D at right-angles to BC [Prop. 1.11], and let the (straight-line) AF have been drawn from A perpendicular to DE [Prop. 1.12], and let GH have been drawn through point F, parallel to BC [Prop. 1.31].
And since BC is at right-angles to each of DA and DE, BC is thus also at right-angles to the plane through ED and DA [Prop. 11.4].
And GH is parallel to it.
And if two straight-lines are parallel, and one of them is at right-angles to some plane, then the remaining (straight-line) will also be at right-angles to the same plane [Prop. 11.8].
Thus, GH is also at right-angles to the plane through ED and DA.
And GH is thus at right-angles to all of the straight-lines joined to it which are also in the plane through ED and AD [Def. 11.3].
And AF, which is in the plane through ED and DA, is joined to it.
Thus, GH is at right-angles to FA.
Hence, FA is also at right-angles to HG.
And AF is also at right-angles to DE.
Thus, AF is at right-angles to each of GH and DE.
And if a straight-line is set up at right-angles to two straight-lines cutting one another, at the point of section, then it will also be at right-angles to the plane through them [Prop. 11.4].
Thus, FA is at right-angles to the plane through ED and GH.
And the plane through ED and GH is the reference (plane).
Thus, AF is at right-angles to the reference plane.
Thus, the straight-line AF has been drawn from the given raised point A perpendicular to the reference plane.
(Which is) the very thing it was required to do.
Proposition 12
To set up a straight-line at right-angles to a given plane from a given point in it.
Let the given plane be the reference (plane), and A a point in it.
So, it is required to set up a straight-line at right-angles to the reference plane at point A.
Let some raised point B have been assumed, and let the perpendicular (straight-line) BC have been drawn from B to the reference plane [Prop. 11.11].
And let AD have been drawn from point A parallel to BC [Prop. 1.31].
Therefore, since AD and CB are two parallel straight-lines, and one of them, BC, is at right-angles to the reference plane, the remaining (one) AD is thus also at right-angles to the reference plane [Prop. 11.8].
Thus, AD has been set up at right-angles to the given plane, from the point in it, A.
(Which is) the very thing it was required to do.
Proposition 13
Two (different) straight-lines cannot be set up at the same point at right-angles to the same plane, on the same side.
For, if possible, let the two straight-lines AB and AC have been set up at the same point A at right-angles to the reference plane, on the same side.
And let the plane through BA and AC have been drawn.
So it will make a straight cutting (passing) through (point) A in the reference plane [Prop. 11.3].
Let it have made DAE.
Thus, AB, AC, and DAE are straight-lines in one plane.
And since CA is at right-angles to the reference plane, it will thus also make right-angles with all of the straight-lines joined to it which are also in the reference plane [Def. 11.3].
And DAE, which is in the reference plane, is joined to it.
Thus, angle CAE is a right-angle.
So, for the same (reasons), BAE is also a right-angle.
Thus, CAE (is) equal to BAE.
And they are in one plane.
The very thing is impossible.
Thus, two (different) straight-lines cannot be set up at the same point at right-angles to the same plane, on the same side.
(Which is) the very thing it was required to show.
Proposition 14
Planes to which the same straight-line is at right-angles will be parallel planes.
For let some straight-line AB be at right-angles to each of the planes CD and EF.
I say that the planes are parallel.
For, if not, being produced, they will meet.
Let them have met.
So they will make a straight-line as a common section [Prop. 11.3].
Let them have made GH.
And let some random point K have been taken on GH.
And let AK and BK have been joined.
And since AB is at right-angles to the plane EF, AB is thus also at right-angles to BK, which is a straight-line in the produced plane EF [Def. 11.3].
Thus, angle ABK is a right-angle.
So, for the same (reasons), BAK is also a right-angle.
So the (sum of the) two angles ABK and BAK in the triangle ABK is equal to two right-angles.
The very thing is impossible [Prop. 1.17].
Thus, planes CD and EF, being produced, will not meet.
Planes CD and EF are thus parallel [Def. 11.8].
Thus, planes to which the same straight-line is at right-angles are parallel planes.
(Which is) the very thing it was required to show.
Proposition 15
If two straight-lines joined to one another are parallel (respectively) to two straight-lines joined to one another, which are not in the same plane, then the planes through them are parallel (to one another).
For let the two straight-lines joined to one another, AB and BC, be parallel to the two straight-lines joined to one another, DE and EF (respectively), not being in the same plane.
I say that the planes through AB, BC and DE, EF will not meet one another (when) produced.
For let BG have been drawn from point B perpendicular to the plane through DE and EF [Prop. 11.11], and let it meet the plane at point G.
And let GH have been drawn through G parallel to ED, and GK (parallel) to EF [Prop. 1.31].
And since BG is at right-angles to the plane through DE and EF, it will thus also make right-angles with all of the straight-lines joined to it, which are also in the plane through DE and EF [Def. 11.3].
And each of GH and GK, which are in the plane through DE and EF, are joined to it.
Thus, each of the angles BGH and BGK are right-angles.
And since BA is parallel to GH [Prop. 11.9], the (sum of the) angles GBA and BGH is equal to two right-angles [Prop. 1.29].
And BGH (is) a right-angle.
GBA (is) thus also a right-angle.
Thus, GB is at right-angles to BA.
So, for the same (reasons), GB is also at right-angles to BC.
Therefore, since the straight-line GB has been set up at right-angles to two straight-lines, BA and BC, cutting one another, GB is thus at right-angles to the plane through BA and BC [Prop. 11.4].
And planes to which the same straight-line is at right-angles are parallel planes [Prop. 11.14].
Thus, the plane through AB and BC is parallel to the (plane) through DE and EF.
Thus, if two straight-lines joined to one another are parallel (respectively) to two straight-lines joined to one another, which are not in the same plane, then the planes through them are parallel (to one another).
(Which is) the very thing it was required to show.
Proposition 16
If two parallel planes are cut by some plane then their common sections are parallel.
For let the two parallel planes AB and CD have been cut by the plane EFHG.
And let EF and GH be their common sections.
I say that EF is parallel to GH.
For, if not, being produced, EF and GH will meet either in the direction of F, H, or of E, G.
Let them be produced, as in the direction of F, H, and let them, first of all, have met at K.
And since EFK is in the plane AB, all of the points on EFK are thus also in the plane AB [Prop. 11.1].
And K is one of the points on EFK.
Thus, K is in the plane AB.
So, for the same (reasons), K is also in the plane CD.
Thus, the planes AB and CD, being produced, will meet.
But they do not meet, on account of being (initially) assumed (to be mutually) parallel.
Thus, the straight-lines EF and GH, being produced in the direction of F, H, will not meet.
So, similarly, we can show that the straight-lines EF and GH, being produced in the direction of E, G, will not meet either.
And (straight-lines in one plane which), being produced, do not meet in either direction are parallel [Def. 1.23].
EF is thus parallel to GH.
Thus, if two parallel planes are cut by some plane then their common sections are parallel.
(Which is) the very thing it was required to show.
Proposition 17
If two straight-lines are cut by parallel planes then they will be cut in the same ratios.
For let the two straight-lines AB and CD be cut by the parallel planes GH, KL, and MN at the points A, E, B, and C, F, D (respectively).
I say that as the straight-line AE is to EB, so CF (is) to FD.
For let AC, BD, and AD have been joined, and let AD meet the plane KL at point O, and let EO and OF have been joined.
And since two parallel planes KL and MN are cut by the plane EBDO, their common sections EO and BD are parallel [Prop. 11.16].
So, for the same (reasons), since two parallel planes GH and KL are cut by the plane AOFC, their common sections AC and OF are parallel [Prop. 11.16].
And since the straight-line EO has been drawn parallel to one of the sides BD of triangle ABD, thus, proportionally, as AE is to EB, so AO (is) to OD [Prop. 6.2].
Again, since the straight-line OF has been drawn parallel to one of the sides AC of triangle ADC, proportionally, as AO is to OD, so CF (is) to FD [Prop. 6.2].
And it was also shown that as AO (is) to OD, so AE (is) to EB.
And thus as AE (is) to EB, so CF (is) to FD [Prop. 5.11].
Thus, if two straight-lines are cut by parallel planes then they will be cut in the same ratios.
(Which is) the very thing it was required to show.
Proposition 18
If a straight-line is at right-angles to some plane then all of the planes (passing) through it will also be at right-angles to the same plane.
For let some straight-line AB be at right-angles to a reference plane.
I say that all of the planes (passing) through AB are also at right-angles to the reference plane.
For let the plane DE have been produced through AB.
And let CE be the common section of the plane DE and the reference (plane).
And let some random point F have been taken on CE.
And let FG have been drawn from F, at right-angles to CE, in the plane DE [Prop. 1.11].
And since AB is at right-angles to the reference plane, AB is thus also at right-angles to all of the straight-lines joined to it which are also in the reference plane [Def. 11.3].
Hence, it is also at right-angles to CE.
Thus, angle ABF is a right-angle.
And GFB is also a right-angle.
Thus, AB is parallel to FG [Prop. 1.28].
And AB is at right-angles to the reference plane.
Thus, FG is also at right-angles to the reference plane [Prop. 11.8].
And a plane is at right-angles to a(nother) plane when the straight-lines drawn at right-angles to the common section of the planes, (and lying) in one of the planes, are at right-angles to the remaining plane [Def. 11.4].
And FG, (which was) drawn at right-angles to the common section of the planes, CE, in one of the planes, DE, was shown to be at right-angles to the reference plane.
Thus, plane DE is at right-angles to the reference (plane).
So, similarly, it can be shown that all of the planes (passing) at random through AB (are) at right-angles to the reference plane.
Thus, if a straight-line is at right-angles to some plane then all of the planes (passing) through it will also be at right-angles to the same plane.
(Which is) the very thing it was required to show.
Proposition 19
If two planes cutting one another are at right-angles to some plane then their common section will also be at right-angles to the same plane.
For let the two planes AB and BC be at right-angles to a reference plane, and let their common section be BD.
I say that BD is at right-angles to the reference plane.
For (if) not, let DE also have been drawn from point D, in the plane AB, at right-angles to the straight-line AD, and DF, in the plane BC, at right-angles to CD.
And since the plane AB is at right-angles to the reference (plane), and DE has been drawn at right-angles to their common section AD, in the plane AB, DE is thus at right-angles to the reference plane [Def. 11.4].
So, similarly, we can show that DF is also at right-angles to the reference plane.
Thus, two (different) straight-lines are set up, at the same point D, at right-angles to the reference plane, on the same side.
The very thing is impossible [Prop. 11.13].
Thus, no (other straight-line) except the common section DB of the planes AB and BC can be set up at point D, at right-angles to the reference plane.
Thus, if two planes cutting one another are at right-angles to some plane then their common section will also be at right-angles to the same plane.
(Which is) the very thing it was required to show.
Proposition 20
If a solid angle is contained by three plane angles then (the sum of) any two (angles) is greater than the remaining (one), (the angles) being taken up in any (possible way).
For let the solid angle A have been contained by the three plane angles BAC, CAD, and DAB.
I say that (the sum of) any two of the angles BAC, CAD, and DAB is greater than the remaining (one), (the angles) being taken up in any (possible way).
For if the angles BAC, CAD, and DAB are equal to one another then (it is) clear that (the sum of) any two is greater than the remaining (one).
But, if not, let BAC be greater (than CAD or DAB).
And let (angle) BAE, equal to the angle DAB, have been constructed in the plane through BAC, on the straight-line AB, at the point A on it.
And let AE be made equal to AD. And BEC being drawn across through point E, let it cut the straight-lines AB and AC at points B and C (respectively).
And let DB and DC have been joined.
And since DA is equal to AE, and AB (is) common, the two (straight-lines AD and AB are) equal to the two (straight-lines EA and AB, respectively).
And angle DAB (is) equal to angle BAE.
Thus, the base DB is equal to the base BE [Prop. 1.4].
And since the (sum of the) two (straight-lines) BD and DC is greater than BC [Prop. 1.20], of which DB was shown (to be) equal to BE, the remainder DC is thus greater than the remainder EC.
And since DA is equal to AE, but AC (is) common, and the base DC is greater than the base EC, the angle DAC is thus greater than the angle EAC [Prop. 1.25].
And DAB was also shown (to be) equal to BAE.
Thus, (the sum of) DAB and DAC is greater than BAC.
So, similarly, we can also show that the remaining (angles), being taken in pairs, are greater than the remaining (one).
Thus, if a solid angle is contained by three plane angles then (the sum of) any two (angles) is greater than the remaining (one), (the angles) being taken up in any (possible way).
(Which is) the very thing it was required to show.
Proposition 21
Any solid angle is contained by plane angles (whose sum is) less [than] four right-angles.
Let the solid angle A be contained by the plane angles BAC, CAD, and DAB.
I say that (the sum of) BAC, CAD, and DAB is less than four right-angles.
For let the random points B, C, and D have been taken on each of (the straight-lines) AB, AC, and AD (respectively).
And let BC, CD, and DB have been joined.
And since the solid angle at B is contained by the three plane angles CBA, ABD, and CBD, (the sum of) any two is greater than the remaining (one) [Prop. 11.20].
Thus, (the sum of) CBA and ABD is greater than CBD.
So, for the same (reasons), (the sum of) BCA and ACD is also greater than BCD, and (the sum of) CDA and ADB is greater than CDB.
Thus, the (sum of the) six angles CBA, ABD, BCA, ACD, CDA, and ADB is greater than the (sum of the) three (angles) CBD, BCD, and CDB.
But, the (sum of the) three (angles) CBD, BDC, and BCD is equal to two right-angles [Prop. 1.32].
Thus, the (sum of the) six angles CBA, ABD, BCA, ACD, CDA, and ADB is greater than two right-angles.
And since the (sum of the) three angles of each of the triangles ABC, ACD, and ADB is equal to two right-angles, the (sum of the) nine angles CBA, ACB, BAC, ACD, CDA, CAD, ADB, DBA, and BAD of the three triangles is equal to six right-angles, of which the (sum of the) six angles ABC, BCA, ACD, CDA, ADB, and DBA is greater than two right-angles.
Thus, the (sum of the) remaining three [angles] BAC, CAD, and DAB, containing the solid angle, is less than four right-angles.
Thus, any solid angle is contained by plane angles (whose sum is) less [than] four right-angles. (Which is) the very thing it was required to show.
Proposition 22
If there are three plane angles, of which (the sum of any) two is greater than the remaining (one), (the angles) being taken up in any (possible way), and if equal straight-lines contain them, then it is possible to construct a triangle from (the straight-lines created by) joining the (ends of the) equal straight-lines.
Let ABC, DEF, and GHK be three plane angles, of which the sum of any) two is greater than the remaining (one), (the angles) being taken up in any (possible way)—(that is), ABC and DEF (greater) than GHK, DEF and GHK (greater) than ABC, and, further, GHK and ABC (greater) than DEF.
And let AB, BC, DE, EF, GH, and HK be equal straight-lines.
And let AC, DF, and GK have been joined.
I say that that it is possible to construct a triangle out of (straight-lines) equal to AC, DF, and GK —that is to say, that (the sum of) any two of AC, DF, and GK is greater than the remaining (one).
Now, if the angles ABC, DEF, and GHK are equal to one another then (it is) clear that, (with) AC, DF, and GK also becoming equal, it is possible to construct a triangle from (straight-lines) equal to AC, DF, and GK.
And if not, let them be unequal, and let KHL, equal to angle ABC, have been constructed on the straight-line HK, at the point H on it.
And let HL be made equal to one of AB, BC, DE, EF, GH, and HK.
And let KL and GL have been joined.
And since the two (straight-lines) AB and BC are equal to the two (straight-lines) KH and HL (respectively), and the angle at B (is) equal to KHL, the base AC is thus equal to the base KL [Prop. 1.4].
And since (the sum of) ABC and GHK is greater than DEF, and ABC equal to KHL, GHL is thus greater than DEF.
And since the two (straight-lines) GH and HL are equal to the two (straight-lines) DE and EF (respectively), and angle GHL (is) greater than DEF, the base GL is thus greater than the base DF [Prop. 1.24].
But, (the sum of) GK and KL is greater than GL [Prop. 1.20].
Thus, (the sum of) GK and KL is much greater than DF.
And KL (is) equal to AC.
Thus, (the sum of) AC and GK is greater than the remaining (straight-line) DF.
So, similarly, we can show that (the sum of) AC and DF is greater than GK, and, further, that (the sum of) DF and GK is greater than AC.
Thus, it is possible to construct a triangle from (straight-lines) equal to AC, DF, and GK.
(Which is) the very thing it was required to show.
Proposition 23
To construct a solid angle from three (given) plane angles, (the sum of) two of which is greater than the remaining (one, the angles) being taken up in any (possible way).
So, it is necessary for the (sum of the) three (angles) to be less than four right-angles [Prop. 11.21].
Let ABC, DEF, and GHK be the three given plane angles, of which let (the sum of) two be greater than the remaining (one, the angles) being taken up in any (possible way), and, further, (let) the (sum of the) three (be) less than four right-angles.
So, it is necessary to construct a solid angle from (plane angles) equal to ABC, DEF, and GHK.
Let AB, BC, DE, EF, GH, and HK be cut off (so as to be) equal (to one another).
And let AC, DF, and GK have been joined.
It is, thus, possible to construct a triangle from (straight-lines) equal to AC, DF, and GK [Prop. 11.22].
Let (such a triangle), LMN, have be constructed, such that AC is equal to LM, DF to MN, and, further, GK to NL.
And let the circle LMN have been circumscribed about triangle LMN [Prop. 4.5].
And let its center have been found, and let it be (at) O.
And let LO, MO, and NO have been joined.
I say that AB is greater than LO.
For, if not, AB is either equal to, or less than, LO.
Let it, first of all, be equal.
And since AB is equal to LO, but AB is equal to BC, and OL to OM, so the two (straight-lines) AB and BC are equal to the two (straight-lines) LO and OM, respectively.
And the base AC was assumed (to be) equal to the base LM.
Thus, angle ABC is equal to angle LOM [Prop. 1.8].
So, for the same (reasons), DEF is also equal to MON, and, further, GHK to NOL.
Thus, the three angles ABC, DEF, and GHK are equal to the three angles LOM, MON, and NOL, respectively.
But, the (sum of the) three angles LOM, MON, and NOL is equal to four right-angles.
Thus, the (sum of the) three angles ABC, DEF, and GHK is also equal to four right-angles.
And it was also assumed (to be) less than four right-angles.
The very thing (is) absurd.
Thus, AB is not equal to LO.
So, I say that AB is not less than LO either.
For, if possible, let it be (less).
And let OP be made equal to AB, and OQ equal to BC, and let PQ have been joined.
And since AB is equal to BC, OP is also equal to OQ.
Hence, the remainder LP is also equal to (the remainder) QM.
LM is thus parallel to PQ [Prop. 6.2], and (triangle) LMO (is) equiangular with (triangle) PQO [Prop. 1.29].
Thus, as OL is to LM, so OP (is) to PQ [Prop. 6.4].
Alternately, as LO (is) to OP, so LM (is) to PQ [Prop. 5.16].
And LO (is) greater than OP.
Thus, LM (is) also greater than PQ [Prop. 5.14].
But LM was made equal to AC.
Thus, AC is also greater than PQ.
Therefore, since the two (straight-lines) AB and BC are equal to the two (straight-lines) PO and OQ (respectively), and the base AC is greater than the base PQ, the angle ABC is thus greater than the angle POQ [Prop. 1.25].
So, similarly, we can show that DEF is also greater than MON, and GHK than NOL.
Thus, the (sum of the) three angles ABC, DEF, and GHK is greater than the (sum of the) three angles LOM, MON, and NOL.
But, (the sum of) ABC, DEF, and GHK was assumed (to be) less than four right-angles.
Thus, (the sum of) LOM, MON, and NOL is much less than four right-angles.
But, (it is) also equal (to four right-angles).
The very thing is absurd.
Thus, AB is not less than LO.
And it was shown (to be) not equal either.
Thus, AB (is) greater than LO.
So let OR have been set up at point O at right-angles to the plane of circle LMN [Prop. 11.12].
And let the (square) on OR be equal to that (area) by which the square on AB is greater than the (square) on LO [Prop. 11.23 lem.].
And let RL, RM, and RN have been joined.
And since RO is at right-angles to the plane of circle LMN, RO is thus also at right-angles to each of LO, MO, and NO.
And since LO is equal to OM, and OR is common and at right-angles, the base RL is thus equal to the base RM [Prop. 1.4].
So, for the same (reasons), RN is also equal to each of RL and RM.
Thus, the three (straight-lines) RL, RM, and RN are equal to one another.
And since the (square) on OR was assumed to be equal to that (area) by which the (square) on AB is greater than the (square) on LO, the (square) on AB is thus equal to the (sum of the squares) on LO and OR.
And the (square) on LR is equal to the (sum of the squares) on LO and OR.
For LOR (is) a right-angle [Prop. 1.47].
Thus, the (square) on AB is equal to the (square) on RL.
Thus, AB (is) equal to RL.
But, each of BC, DE, EF, GH, and HK is equal to AB, and each of RM and RN equal to RL.
Thus, each of AB, BC, DE, EF, GH, and HK is equal to each of RL, RM, and RN.
And since the two (straight-lines) LR and RM are equal to the two (straight-lines) AB and BC (respectively), and the base LM was assumed (to be) equal to the base AC, the angle LRM is thus equal to the angle ABC [Prop. 1.8].
So, for the same (reasons), MRN is also equal to DEF, and LRN to GHK.
Thus, the solid angle R, contained by the angles LRM, MRN, and LRN, has been constructed out of the three plane angles LRM, MRN, and LRN, which are equal to the three given (plane angles) ABC, DEF, and GHK (respectively).
(Which is) the very thing it was required to do.
Lemma
And we can demonstrate, thusly, in which manner to take the (square) on OR equal to that (area) by which the (square) on AB is greater than the (square) on LO.
Let the straight-lines AB and LO be set out, and let AB be greater, and let the semicircle ACB have been drawn around it.
And let AC, equal to the straight-line LO, which is not greater than the diameter AB, have been inserted into the semicircle ACB [Prop. 4.1].
And let CB have been joined.
Therefore, since the angle ACB is in the semicircle ACB, ACB is thus a right-angle [Prop. 3.31].
Thus, the (square) on AB is equal to the (sum of the) squares on AC and CB [Prop. 1.47].
Hence, the (square) on AB is greater than the (square) on AC by the (square) on CB.
And AC (is) equal to LO.
Thus, the (square) on AB is greater than the (square) on LO by the (square) on CB.
Therefore, if we take OR equal to BC then the (square) on AB will be greater than the (square) on LO by the (square) on OR.
(Which is) the very thing it was prescribed to do.
Proposition 24
If a solid (figure) is contained by (six) parallel planes then its opposite planes are both equal and parallelogrammic.
For let the solid (figure) CDHG have been contained by the parallel planes AC, GF, and AH, DF, and BF, AE.
I say that its opposite planes are both equal and parallelogrammic.
For since the two parallel planes BG and CE are cut by the plane AC, their common sections are parallel [Prop. 11.16].
Thus, AB is parallel to DC.
Again, since the two parallel planes BF and AE are cut by the plane AC, their common sections are parallel [Prop. 11.16].
Thus, BC is parallel to AD.
And AB was also shown (to be) parallel to DC.
Thus, AC is a parallelogram.
So, similarly, we can also show that DF, FG, GB, BF, and AE are each parallelograms.
Let AH and DF have been joined.
And since AB is parallel to DC, and BH to CF, so the two (straight-lines) joining one another, AB and BH, are parallel to the two straight-lines joining one another, DC and CF (respectively), not (being) in the same plane.
Thus, they will contain equal angles [Prop. 11.10].
Thus, angle ABH (is) equal to (angle) DCF.
And since the two (straight-lines) AB and BH are equal to the two (straight-lines) DC and CF (respectively) [Prop. 1.34], and angle ABH is equal to angle DCF, the base AH is thus equal to the base DF, and triangle ABH is equal to triangle DCF [Prop. 1.4].
And parallelogram BG is double (triangle) ABH, and parallelogram CE double (triangle) DCF [Prop. 1.34].
Thus, parallelogram BG (is) equal to parallelogram CE.
So, similarly, we can show that AC is also equal to GF, and AE to BF.
Thus, if a solid (figure) is contained by (six) parallel planes then its opposite planes are both equal and parallelogrammic.
(Which is) the very thing it was required to show.
Proposition 25
If a parallelipiped solid is cut by a plane which is parallel to the opposite planes (of the parallelipiped) then as the base (is) to the base, so the solid will be to the solid.
For let the parallelipiped solid ABCD have been cut by the plane FG which is parallel to the opposite planes RA and DH.
I say that as the base AEFV (is) to the base EHCF, so the solid ABFU (is) to the solid EGCD.
For let AH have been produced in each direction.
And let any number whatsoever (of lengths), AK and KL, be made equal to AE, and any number whatsoever (of lengths), HM and MN, equal to EH.
And let the parallelograms LP, KV, HW, and MS have been completed, and the solids LQ, KR, DM, and MT.
And since the straight-lines LK, KA, and AE are equal to one another, the parallelograms LP, KV, and AF are also equal to one another, and KO, KB, and AG (are equal) to one another, and, further, LX, KQ, and AR (are equal) to one another.
For (they are) opposite [Prop. 11.24].
So, for the same (reasons), the parallelograms EC, HW, and MS are also equal to one another, and HG, HI, and IN are equal to one another, and, further, DH, MY, and NT (are equal to one another).
Thus, three planes of (one of) the solids LQ, KR, and AU are equal to the (corresponding) three planes (of the others).
But, the three planes (in one of the soilds) are equal to the three opposite planes [Prop. 11.24].
Thus, the three solids LQ, KR, and AU are equal to one another [Def. 11.10].
So, for the same (reasons), the three solids ED, DM, and MT are also equal to one another.
Thus, as many multiples as the base LF is of the base AF, so many multiples is the solid LU also of the the solid AU.
So, for the same (reasons), as many multiples as the base NF is of the base FH, so many multiples is the solid NU also of the solid HU.
And if the base LF is equal to the base NF then the solid LU is also equal to the solid NU.
And if the base LF exceeds the base NF then the solid LU also exceeds the solid NU.
And if ( LF) is less than ( NF) then ( LU) is (also) less than ( NU).
So, there are four magnitudes, the two bases AF and FH, and the two solids AU and UH, and equal multiples have been taken of the base AF and the solid AU ---(namely), the base LF and the solid LU ---and of the base HF and the solid HU ---(namely), the base NF and the solid NU.
And it has been shown that if the base LF exceeds the base FN then the solid LU also exceeds the [solid] NU, and if ( LF is) equal (to FN) then ( LU is) equal (to NU), and if ( LF is) less than ( FN) then ( LU is) less than ( NU).
Thus, as the base AF is to the base FH, so the solid AU (is) to the solid UH [Def. 5.5].
(Which is) the very thing it was required to show.
Proposition 26
To construct a solid angle equal to a given solid angle on a given straight-line, and at a given point on it.
Let AB be the given straight-line, and A the given point on it, and D the given solid angle, contained by the plane angles EDC, EDF, and FDC.
So, it is necessary to construct a solid angle equal to the solid angle D on the straight-line AB, and at the point A on it.
For let some random point F have been taken on DF, and let FG have been drawn from F perpendicular to the plane through ED and DC [Prop. 11.11], and let it meet the plane at G, and let DG have been joined.
And let BAL, equal to the angle EDC, and BAK, equal to EDG, have been constructed on the straight-line AB at the point A on it [Prop. 1.23].
And let AK be made equal to DG.
And let KH have been set up at the point K at right-angles to the plane through BA and AL [Prop. 11.12].
And let KH be made equal to GF.
And let HA have been joined.
I say that the solid angle at A, contained by the (plane) angles BAL, BAH, and HAL, is equal to the solid angle at D, contained by the (plane) angles EDC, EDF, and FDC.
For let AB and DE have been cut off (so as to be) equal, and let HB, KB, FE, and GE have been joined.
And since FG is at right-angles to the reference plane (through ED and DC), it will also make right-angles with all of the straight-lines joined to it which are also in the reference plane [Def. 11.3].
Thus, the angles FGD and FGE are right-angles.
So, for the same (reasons), the angles HKA and HKB are also right-angles.
And since the two (straight-lines) KA and AB are equal to the two (straight-lines) GD and DE, respectively, and they contain equal angles, the base KB is thus equal to the base GE [Prop. 1.4].
And KH is also equal to GF.
And they contain right-angles (with the respective bases).
Thus, HB (is) also equal to FE [Prop. 1.4].
Again, since the two (straight-lines) AK and KH are equal to the two (straight-lines) DG and GF (respectively), and they contain right-angles, the base AH is thus equal to the base FD [Prop. 1.4].
And AB (is) also equal to DE.
So, the two (straight-lines) HA and AB are equal to the two (straight-lines) DF and DE (respectively).
And the base HB (is) equal to the base FE.
Thus, the angle BAH is equal to the angle EDF [Prop. 1.8].
So, for the same (reasons), HAL is also equal to FDC.
And BAL is also equal to EDC.
Thus, (a solid angle) has been constructed, equal to the given solid angle at D, on the given straight-line AB, at the given point A on it.
(Which is) the very thing it was required to do.
Proposition 27
To describe a parallelepiped solid similar, and similarly laid out, to a given parallelepiped solid on a given straight-line.
Let the given straight-line be AB, and the given parallelepiped solid CD.
So, it is necessary to describe a parallelepiped solid similar, and similarly laid out, to the given parallelepiped solid CD on the given straight-line AB.
For, let a (solid angle) contained by the (plane angles) BAH, HAK, and KAB have been constructed, equal to solid angle at C, on the straight-line AB at the point A on it [Prop. 11.26], such that angle BAH is equal to ECF, and BAK to ECG, and KAH to GCF.
And let it have been contrived that as EC (is) to CG, so BA (is) to AK, and as GC (is) to CF, so KA (is) to AH [Prop. 6.12].
And thus, via equality, as EC is to CF, so BA (is) to AH [Prop. 5.22].
And let the parallelogram HB have been completed, and the solid AL.
And since as EC is to CG, so BA (is) to AK, and the sides about the equal angles ECG and BAK are (thus) proportional, the parallelogram GE is thus similar to the parallelogram KB.
So, for the same (reasons), the parallelogram KH is also similar to the parallelogram GF, and, further, FE (is similar) to HB.
Thus, three of the parallelograms of solid CD are similar to three of the parallelograms of solid AL.
But, the (former) three are equal and similar to the three opposite, and the (latter) three are equal and similar to the three opposite.
Thus, the whole solid CD is similar to the whole solid AL [Def. 11.9].
Thus, AL, similar, and similarly laid out, to the given parallelepiped solid CD, has been described on the given straight-lines AB.
(Which is) the very thing it was required to do.
Proposition 28
If a parallelepiped solid is cut by a plane (passing) through the diagonals of (a pair of) opposite planes then the solid will be cut in half by the plane.
For let the parallelepiped solid AB have been cut by the plane CDEF (passing) through the diagonals of the opposite planes CF and DE.
I say that the solid AB will be cut in half by the plane CDEF.
For since triangle CGF is equal to triangle CFB, and ADE (is equal) to DEH [Prop. 1.34], and parallelogram CA is also equal to EB ---for (they are) opposite [Prop. 11.24] ---and GE (equal) to CH, thus the prism contained by the two triangles CGF and ADE, and the three parallelograms GE, AC, and CE, is also equal to the prism contained by the two triangles CFB and DEH, and the three parallelograms CH, BE, and CE.
For they are contained by planes (which are) equal in number and in magnitude [Def. 11.10].
Thus, the whole of solid AB is cut in half by the plane CDEF.
(Which is) the very thing it was required to show.
Proposition 29
Parallelepiped solids which are on the same base, and (have) the same height, and in which the (ends of the straight-lines) standing up are on the same straight-lines, are equal to one another.
For let the parallelepiped solids CM and CN be on the same base AB, and (have) the same height, and let the (ends of the straight-lines) standing up in them, AG, AF, LM, LN, CD, CE, BH, and BK, be on the same straight-lines, FN and DK.
I say that solid CM is equal to solid CN.
For since CH and CK are each parallelograms, CB is equal to each of DH and EK [Prop. 1.34].
Hence, DH is also equal to EK.
Let EH have been subtracted from both.
Thus, the remainder DE is equal to the remainder HK.
Hence, triangle DCE is also equal to triangle HBK [Prop. 1.4] [Prop. 1.8], and parallelogram DG to parallelogram HN [Prop. 1.36].
So, for the same (reasons), traingle AFG is also equal to triangle MLN.
And parallelogram CF is also equal to parallelogram BM, and CG to BN [Prop. 11.24].
For they are opposite.
Thus, the prism contained by the two triangles AFG and DCE, and the three parallelograms AD, DG, and CG, is equal to the prism contained by the two triangles MLN and HBK, and the three parallelograms BM, HN, and BN.
Let the solid whose base (is) parallelogram AB, and (whose) opposite (face is) GEHM, have been added to both (prisms).
Thus, the whole parallelepiped solid CM is equal to the whole parallelepiped solid CN.
Thus, parallelepiped solids which are on the same base, and (have) the same height, and in which the (ends of the straight-lines) standing up (are) on the same straight-lines, are equal to one another.
(Which is) the very thing it was required to show.
Proposition 30
Parallelepiped solids which are on the same base, and (have) the same height, and in which the (ends of the straight-lines) standing up are not on the same straight-lines, are equal to one another.
Let the parallelepiped solids CM and CN be on the same base, AB, and (have) the same height, and let the (ends of the straight-lines) standing up in them, AF, AG, LM, LN, CD, CE, BH, and BK, not be on the same straight-lines.
I say that the solid CM is equal to the solid CN.
For let NK and DH have been produced, and let them have joined one another at R.
And, further, let FM and GE have been produced to P and Q (respectively).
And let AO, LP, CQ, and BR have been joined.
So, solid CM, whose base (is) parallelogram ACBL, and opposite (face) FDHM, is equal to solid CP, whose base (is) parallelogram ACBL, and opposite (face) OQRP.
For they are on the same base, ACBL, and (have) the same height, and the (ends of the straight-lines) standing up in them, AF, AO, LM, LP, CD, CQ, BH, and BR, are on the same straight-lines, FP and DR [Prop. 11.29].
But, solid CP, whose base is parallelogram ACBL, and opposite (face) OQRP, is equal to solid CN, whose base (is) parallelogram ACBL, and opposite (face) GEKN.
For, again, they are on the same base, ACBL, and (have) the same height, and the (ends of the straight-lines) standing up in them, AG, AO, CE, CQ, LN, LP, BK, and BR, are on the same straight-lines, GQ and NR [Prop. 11.29].
Hence, solid CM is also equal to solid CN.
Thus, parallelepiped solids (which are) on the same base, and (have) the same height, and in which the (ends of the straight-lines) standing up are not on the same straight-lines, are equal to one another.
(Which is) the very thing it was required to show.
Proposition 31
Parallelepiped solids which are on equal bases, and (have) the same height, are equal to one another.
Let the parallelepiped solids AE and CF be on the equal bases AB and CD (respectively), and (have) the same height.
I say that solid AE is equal to solid CF.
So, let the (straight-lines) standing up, HK, BE, AG, LM, PQ, DF, CO, and RS, first of all, be at right-angles to the bases AB and CD.
And let RT have been produced in a straight-line with CR.
And let (angle) TRU, equal to angle ALB, have been constructed on the straight-line RT, at the point R on it [Prop. 1.23].
And let RT be made equal to AL, and RU to LB.
And let the base RW, and the solid XU, have been completed.
And since the two (straight-lines) TR and RU are equal to the two (straight-lines) AL and LB (respectively), and they contain equal angles, parallelogram RW is thus equal and similar to parallelogram HL [Prop. 6.14].
And, again, since AL is equal to RT, and LM to RS, and they contain right-angles, parallelogram RX is thus equal and similar to parallelogram AM [Prop. 6.14].
So, for the same (reasons), LE is also equal and similar to SU.
Thus, three parallelograms of solid AE are equal and similar to three parallelograms of solid XU.
But, the three (faces of the former solid) are equal and similar to the three opposite (faces), and the three (faces of the latter solid) to the three opposite (faces) [Prop. 11.24].
Thus, the whole parallelepiped solid AE is equal to the whole parallelepiped solid XU [Def. 11.10].
Let DR and WU have been drawn across, and let them have met one another at Y.
And let aTb have been drawn through T parallel to DY.
And let PD have been produced to a.
And let the solids YX and RI have been completed.
So, solid XY, whose base is parallelogram RX, and opposite (face) Yc, is equal to solid XU, whose base (is) parallelogram RX, and opposite (face) UV.
For they are on the same base RX, and (have) the same height, and the (ends of the straight-lines) standing up in them, RY, RU, Tb, TW, Se, Sd, Xc and XV, are on the same straight-lines, YW and eV [Prop. 11.29].
But, solid XU is equal to AE.
Thus, solid XY is also equal to solid AE.
And since parallelogram RUWT is equal to parallelogram YT.
For they are on the same base RT, and between the same parallels RT and YW [Prop. 1.35].
But, RUWT is equal to CD, since (it is) also (equal) to AB.
Parallelogram YT is thus also equal to CD.
And DT is another (parallelogram).
Thus, as base CD is to DT, so YT (is) to DT [Prop. 5.7].
And since the parallelepiped solid CI has been cut by the plane RF, which is parallel to the opposite planes (of CI), as base CD is to base DT, so solid CF (is) to solid RI [Prop. 11.25].
So, for the same (reasons), since the parallelepiped solid YI has been cut by the plane RX, which is parallel to the opposite planes (of YI), as base YT is to base TD, so solid YX (is) to solid RI [Prop. 11.25].
But, as base CD (is) to DT, so YT (is) to DT.
And, thus, as solid CF (is) to solid RI, so solid YX (is) to solid RI.
Thus, solids CF and YX each have the same ratio to RI [Prop. 5.11].
Thus, solid CF is equal to solid YX [Prop. 5.9].
But, YX was show (to be) equal to AE.
Thus, AE is also equal to CF.
And so let the (straight-lines) standing up, AG, HK, BE, LM, CO, PQ, DF, and RS, not be at right-angles to the bases AB and CD.
Again, I say that solid AE (is) equal to solid CF.
For let KN, ET, GU, MV, QW, FX, OY, and SI have been drawn from points K, E, G, M, Q, F, O, and S (respectively) perpendicular to the reference plane (i.e., the plane of the bases AB and CD), and let them have met the plane at points N, T, U, V, W, X, Y, and I (respectively).
And let NT, NU, UV, TV, WX, WY, YI, and IX have been joined.
So solid KV is equal to solid QI.
For they are on the equal bases KM and QS, and (have) the same height, and the (straight-lines) standing up in them are at right-angles to their bases (see first part of proposition).
But, solid KV is equal to solid AE, and QI to CF.
For they are on the same base, and (have) the same height, and the (straight-lines) standing up in them are not on the same straight-lines [Prop. 11.30].
Thus, solid AE is also equal to solid CF.
Thus, parallelepiped solids which are on equal bases, and (have) the same height, are equal to one another.
(Which is) the very thing it was required to show.
Proposition 32
Parallelepiped solids which (have) the same height are to one another as their bases.
Let AB and CD be parallelepiped solids (having) the same height.
I say that the parallelepiped solids AB and CD are to one another as their bases.
That is to say, as base AE is to base CF, so solid AB (is) to solid CD.
For let FH, equal to AE, have been applied to FG (in the angle FGH equal to angle LCG) [Prop. 1.45].
And let the parallelepiped solid GK, (having) the same height as CD, have been completed on the base FH.
So solid AB is equal to solid GK.
For they are on the equal bases AE and FH, and (have) the same height [Prop. 11.31].
And since the parallelepiped solid CK has been cut by the plane DG, which is parallel to the opposite planes (of CK), thus as the base CF is to the base FH, so the solid CD (is) to the solid DH [Prop. 11.25].
And base FH (is) equal to base AE, and solid GK to solid AB.
And thus as base AE is to base CF, so solid AB (is) to solid CD.
Thus, parallelepiped solids which (have) the same height are to one another as their bases.
(Which is) the very thing it was required to show.
Proposition 35
If there are two equal plane angles, and raised straight-lines are stood on the apexes of them, containing equal angles respectively with the original straight-lines (forming the angles), and random points are taken on the raised (straight-lines), and perpendiculars are drawn from them to the planes in which the original angles are, and straight-lines are joined from the points created in the planes to the (vertices of the) original angles, then they will enclose equal angles with the raised (straight-lines).
Let BAC and EDF be two equal rectilinear angles.
And let the raised straight-lines AG and DM have been stood on points A and D, containing equal angles respectively with the original straight-lines.
(That is) MDE (equal) to GAB, and MDF (to) GAC.
And let the random points G and M have been taken on AG and DM (respectively).
And let the GL and MN have been drawn from points G and M perpendicular to the planes through BAC and EDF (respectively).
And let them have joined the planes at points L and N (respectively).
And let LA and ND have been joined.
I say that angle GAL is equal to angle MDN.
Let AH be made equal to DM.
And let HK have been drawn through point H parallel to GL.
And GL is perpendicular to the plane through BAC.
Thus, HK is also perpendicular to the plane through BAC [Prop. 11.8].
And let KC, NF, KB, and NE have been drawn from points K and N perpendicular to the straight-lines AC, DF, AB, and DE.
And let HC, CB, MF, and FE have been joined.
Since the (square) on HA is equal to the (sum of the squares) on HK and KA [Prop. 1.47], and the (sum of the squares) on KC and CA is equal to the (square) on KA [Prop. 1.47], thus the (square) on HA is equal to the (sum of the squares) on HK, KC, and CA.
And the (square) on HC is equal to the (sum of the squares) on HK and KC [Prop. 1.47].
Thus, the (square) on HA is equal to the (sum of the squares) on HC and CA.
Thus, angle HCA is a right-angle [Prop. 1.48].
So, for the same (reasons), angle DFM is also a right-angle.
Thus, angle ACH is equal to (angle) DFM.
And HAC is also equal to MDF.
So, MDF and HAC are two triangles having two angles equal to two angles, respectively, and one side equal to one side---(namely), that subtending one of the equal angles ---(that is), HA (equal) to MD.
Thus, they will also have the remaining sides equal to the remaining sides, respectively [Prop. 1.26].
Thus, AC is equal to DF.
So, similarly, we can show that AB is also equal to DE.
Therefore, since AC is equal to DF, and AB to DE, so the two (straight-lines) CA and AB are equal to the two (straight-lines) FD and DE (respectively).
But, angle CAB is also equal to angle FDE.
Thus, base BC is equal to base EF, and triangle ( ACB) to triangle ( DFE), and the remaining angles to the remaining angles (respectively) [Prop. 1.4].
Thus, angle ACB (is) equal to DFE.
And the right-angle ACK is also equal to the right-angle DFN.
Thus, the remainder BCK is equal to the remainder EFN.
So, for the same (reasons), CBK is also equal to FEN.
So, BCK and EFN are two triangles having two angles equal to two angles, respectively, and one side equal to one side—(namely), that by the equal angles—(that is), BC (equal) to EF.
Thus, they will also have the remaining sides equal to the remaining sides (respectively) [Prop. 1.26].
Thus, CK is equal to FN.
And AC (is) also equal to DF.
So, the two (straight-lines) AC and CK are equal to the two (straight-lines) DF and FN (respectively).
And they enclose right-angles.
Thus, base AK is equal to base DN [Prop. 1.4].
And since AH is equal to DM, the (square) on AH is also equal to the (square) on DM.
But, the the (sum of the squares) on AK and KH is equal to the (square) on AH.
For angle AKH (is) a right-angle [Prop. 1.47].
And the (sum of the squares) on DN and NM (is) equal to the square on DM.
For angle DNM (is) a right-angle [Prop. 1.47].
Thus, the (sum of the squares) on AK and KH is equal to the (sum of the squares) on DN and NM, of which the (square) on AK is equal to the (square) on DN.
Thus, the remaining (square) on KH is equal to the (square) on NM.
Thus, HK (is) equal to MN.
And since the two (straight-lines) HA and AK are equal to the two (straight-lines) MD and DN, respectively, and base HK was shown (to be) equal to base MN, angle HAK is thus equal to angle MDN [Prop. 1.8].
Thus, if there are two equal plane angles, and so on of the proposition.
[(Which is) the very thing it was required to show].
Corollary
So, it is clear, from this, that if there are two equal plane angles, and equal raised straight-lines are stood on them (at their apexes), containing equal angles respectively with the original straight-lines (forming the angles), then the perpendiculars drawn from (the raised ends of) them to the planes in which the original angles lie are equal to one another.
(Which is) the very thing it was required to show.
Proposition 36
If three straight-lines are (continuously) proportional then the parallelepiped solid (formed) from the three (straight-lines) is equal to the equilateral parallelepiped solid on the middle (straight-line which is) equiangular to the aforementioned (parallelepiped solid).
Let A, B, and C be three (continuously) proportional straight-lines, (such that) as A (is) to B, so B (is) to C.
I say that the (parallelepiped) solid (formed) from A, B, and C is equal to the equilateral solid on B (which is) equiangular with the aforementioned (solid).
Let the solid angle at E, contained by DEG, GEF, and FED, be set out.
And let DE, GE, and EF each be made equal to B.
And let the parallelepiped solid EK have been completed.
And (let) LM (be made) equal to A.
And let the solid angle contained by NLO, OLM, and MLN have been constructed on the straight-line LM, and at the point L on it, (so as to be) equal to the solid angle E [Prop. 11.23].
And let LO be made equal to B, and LN equal to C.
And since as A (is) to B, so B (is) to C, and A (is) equal to LM, and B to each of LO and ED, and C to LN, thus as LM (is) to EF, so DE (is) to LN.
And (so) the sides around the equal angles NLM and DEF are reciprocally proportional.
Thus, parallelogram MN is equal to parallelogram DF [Prop. 6.14].
And since the two plane rectilinear angles DEF and NLM are equal, and the raised straight-lines stood on them (at their apexes), LO and EG, are equal to one another, and contain equal angles respectively with the original straight-lines (forming the angles), the perpendiculars drawn from points G and O to the planes through NLM and DEF (respectively) are thus equal to one another [Prop. 11.35 corr.].
Thus, the solids LH and EK (have) the same height.
And parallelepiped solids on equal bases, and with the same height, are equal to one another [Prop. 11.31].
Thus, solid HL is equal to solid EK.
And LH is the solid (formed) from A, B, and C, and EK the solid on B.
Thus, the parallelepiped solid (formed) from A, B, and C is equal to the equilateral solid on B (which is) equiangular with the aforementioned (solid).
(Which is) the very thing it was required to show.
Proposition 37
If four straight-lines are proportional then the similar, and similarly described, parallelepiped solids on them will also be proportional.
And if the similar, and similarly described, parallelepiped solids on them are proportional then the straight-lines themselves will be proportional.
Let AB, CD, EF, and GH, be four proportional straight-lines, (such that) as AB (is) to CD, so EF (is) to GH.
And let the similar, and similarly laid out, parallelepiped solids KA, LC, ME and NG have been described on AB, CD, EF, and GH (respectively).
I say that as KA is to LC, so ME (is) to NG.
For since the parallelepiped solid KA is similar to LC, KA thus has to LC the cubed ratio that AB (has) to CD [Prop. 11.33].
So, for the same (reasons), ME also has to NG the cubed ratio that EF (has) to GH [Prop. 11.33].
And since as AB is to CD, so EF (is) to GH, thus, also, as AK (is) to LC, so ME (is) to NG.
And so let solid AK be to solid LC, as solid ME (is) to NG.
I say that as straight-line AB is to CD, so EF (is) to GH.
For, again, since KA has to LC the cubed ratio that AB (has) to CD [Prop. 11.33], and ME also has to NG the cubed ratio that EF (has) to GH [Prop. 11.33], and as KA is to LC, so ME (is) to NG, thus, also, as AB (is) to CD, so EF (is) to GH.
Thus, if four straight-lines are proportional, and so on of the proposition.
(Which is) the very thing it was required to show.
Proposition 38
If the sides of the opposite planes of a cube are cut in half, and planes are produced through the pieces, then the common section of the (latter) planes and the diameter of the cube cut one another in half.
For let the opposite planes CF and AH of the cube AF have been cut in half at the points K, L, M, N, O, Q, P, and R.
And let the planes KN and OR have been produced through the pieces.
And let US be the common section of the planes, and DG the diameter of cube AF.
I say that UT is equal to TS, and DT to TG.
For let DU, UE, BS, and SG have been joined.
And since DO is parallel to PE, the alternate angles DOU and UPE are equal to one another [Prop. 1.29].
And since DO is equal to PE, and OU to UP, and they contain equal angles, base DU is thus equal to base UE, and triangle DOU is equal to triangle PUE, and the remaining angles (are) equal to the remaining angles [Prop. 1.4].
Thus, angle OUD (is) equal to angle PUE.
So, for this (reason), DUE is a straight-line [Prop. 1.14].
So, for the same (reason), BSG is also a straight-line, and BS equal to SG.
And since CA is equal and parallel to DB, but CA is also equal and parallel to EG, DB is thus also equal and parallel to EG [Prop. 11.9].
And the straight-lines DE and BG join them.
DE is thus parallel to BG [Prop. 1.33].
Thus, angle EDT (is) equal to BGT.
For (they are) alternate [Prop. 1.29].
And (angle) DTU (is equal) to GTS [Prop. 1.15].
So, DTU and GTS are two triangles having two angles equal to two angles, and one side equal to one side—(namely), that subtended by one of the equal angles—(that is), DU (equal) to GS.
For they are halves of DE and BG (respectively).
(Thus), they will also have the remaining sides equal to the remaining sides [Prop. 1.26].
Thus, DT (is) equal to TG, and UT to TS.
Thus, if the sides of the opposite planes of a cube are cut in half, and planes are produced through the pieces, then the common section of the (latter) planes and the diameter of the cube cut one another in half.
(Which is) the very thing it was required to show.
Proposition 39
If there are two equal height prisms, and one has a parallelogram, and the other a triangle, (as a) base, and the parallelogram is double the triangle, then the prisms will be equal.
Let ABCDEF and GHKLMN be two equal height prisms, and let the former have the parallelogram AF, and the latter the triangle GHK, as a base.
And let parallelogram AF be twice triangle GHK.
I say that prism ABCDEF is equal to prism GHKLMN.
For let the solids AO and GP have been completed.
Since parallelogram AF is double triangle GHK, and parallelogram HK is also double triangle GHK [Prop. 1.34], parallelogram AF is thus equal to parallelogram HK.
And parallelepiped solids which are on equal bases, and (have) the same height, are equal to one another [Prop. 11.31].
Thus, solid AO is equal to solid GP.
And prism ABCDEF is half of solid AO, and prism GHKLMN half of solid GP [Prop. 11.28].
Prism ABCDEF is thus equal to prism GHKLMN.
Thus, if there are two equal height prisms, and one has a parallelogram, and the other a triangle, (as a) base, and the parallelogram is double the triangle, then the prisms are equal.
(Which is) the very thing it was required to show.