Book 3 — Fundamentals of Plane Geometry Involving Circles ◧
Definitions
1. Equal circles are (circles) whose diameters are equal, or whose (distances) from the centers (to the circumferences) are equal (i.e., whose radii are equal).
2. A straight-line said to touch a circle is any (straight-line) which, meeting the circle and being produced, does not cut the circle.
3. Circles said to touch one another are any (circles) which, meeting one another, do not cut one another.
4. In a circle, straight-lines are said to be equally far from the center when the perpendiculars drawn to them from the center are equal.
5. And (that straight-line) is said to be further (from the center) on which the greater perpendicular falls (from the center).
6. A segment of a circle is the figure contained by a straight-line and a circumference of a circle.
7. And the angle of a segment is that contained by a straight-line and a circumference of a circle.
8. And the angle in a segment is the angle contained by the joined straight-lines, when any point is taken on the circumference of a segment, and straight-lines are joined from it to the ends of the straight-line which is the base of the segment.
9. And when the straight-lines containing an angle cut off some circumference, the angle is said to stand upon that (circumference).
10. And a sector of a circle is the figure contained by the straight-lines surrounding an angle, and the circumference cut off by them, when the angle is constructed at the center of a circle.
11. Similar segments of circles are those accepting equal angles, or in which the angles are equal to one another.
Proposition 1
To find the center of a given circle.
Let ABC be the given circle.
So it is required to find the center of circle ABC.
Let some straight-line AB have been drawn through ( ABC), at random, and let ( AB) have been cut in half at point D [Prop. 1.9].
And let DC have been drawn from D, at right-angles to AB [Prop. 1.11].
And let ( CD) have been drawn through to E (as CE).
And let CE have been cut in half at F [Prop. 1.9].
I say that (point) F is the center of the [circle] ABC.
For (if) not then, if possible, let G (be the center of the circle), and let GA, GD, and GB have been joined.
And since AD is equal to DB, and DG (is) common, the two (straight-lines) AD, DG are equal to the two (straight-lines) BD, DG, respectively.
And the base GA is equal to the base GB.
For (they are both) radii.
Thus, angle ADG is equal to angle GDB [Prop. 1.8].
And when a straight-line stood upon (another) straight-line make adjacent angles (which are) equal to one another, each of the equal angles is a right-angle [Def. 1.10].
Thus, GDB is a right-angle.
And FDB is also a right-angle.
Thus, FDB (is) equal to GDB, the greater to the lesser.
The very thing is impossible.
Thus, (point) G is not the center of the circle ABC.
So, similarly, we can show that neither is any other (point) except F.
Thus, point F is the center of the [circle] ABC.
Corollary
So, from this, (it is) manifest that if any straight-line in a circle cuts any (other) straight-line in half, and at right-angles, then the center of the circle is on the former (straight-line).
(Which is) the very thing it was required to do.
Proposition 2
If two points are taken at random on the circumference of a circle then the straight-line joining the points will fall inside the circle.
Let ABC be a circle, and let two points A and B have been taken at random on its circumference.
I say that the straight-line joining A to B will fall inside the circle.
For (if) not then, if possible, let it fall outside (the circle), like AEB (in the figure).
And let the center of the circle ABC have been found [Prop. 3.1], and let it be (at point) D.
And let DA and DB have been joined, and let DFE have been drawn through.
Therefore, since DA is equal to DB, the angle DAE (is) thus also equal to DBE [Prop. 1.5].
And since in triangle AED the one side, AEB, has been produced, angle DEB (is) thus greater than DAE [Prop. 1.16].
And DAE (is) equal to DBE [Prop. 1.5].
Thus, DEB (is) greater than DBE.
And the greater angle is subtended by the greater side [Prop. 1.19].
Thus, DB (is) greater than DE.
And DB (is) equal to DF.
Thus, DF (is) greater than DE, the lesser than the greater.
The very thing is impossible.
Thus, the straight-line joining A to B will not fall outside the circle.
So, similarly, we can show that neither (will it fall) on the circumference itself.
Thus, (it will fall) inside (the circle).
Thus, if two points are taken at random on the circumference of a circle then the straight-line joining the points will fall inside the circle.
(Which is) the very thing it was required to show.
Proposition 3
In a circle, if any straight-line through the center cuts in half any straight-line not through the center then it also cuts it at right-angles.
And (conversely) if it cuts it at right-angles then it also cuts it in half.
Let ABC be a circle, and, within it, let some straight-line through the center, CD, cut in half some straight-line not through the center, AB, at the point F.
I say that ( CD) also cuts ( AB) at right-angles.
For let the center of the circle ABC have been found [Prop. 3.1], and let it be (at point) E, and let EA and EB have been joined.
And since AF is equal to FB, and FE (is) common, two (sides of triangle AFE) [are] equal to two (sides of triangle BFE).
And the base EA (is) equal to the base EB.
Thus, angle AFE is equal to angle BFE [Prop. 1.8].
And when a straight-line stood upon (another) straight-line makes adjacent angles (which are) equal to one another, each of the equal angles is a right-angle [Def. 1.10].
Thus, AFE and BFE are each right-angles.
Thus, the (straight-line) CD, which is through the center and cuts in half the (straight-line) AB, which is not through the center, also cuts ( AB) at right-angles.
And so let CD cut AB at right-angles.
I say that it also cuts ( AB) in half.
That is to say, that AF is equal to FB.
For, with the same construction, since EA is equal to EB, angle EAF is also equal to EBF [Prop. 1.5].
And the right-angle AFE is also equal to the right-angle BFE.
Thus, EAF and EFB are two triangles having two angles equal to two angles, and one side equal to one side---(namely), their common (side) EF, subtending one of the equal angles.
Thus, they will also have the remaining sides equal to the (corresponding) remaining sides [Prop. 1.26].
Thus, AF (is) equal to FB.
Thus, in a circle, if any straight-line through the center cuts in half any straight-line not through the center then it also cuts it at right-angles.
And (conversely) if it cuts it at right-angles then it also cuts it in half.
(Which is) the very thing it was required to show.
Proposition 4
In a circle, if two straight-lines, which are not through the center, cut one another then they do not cut one another in half.
Let ABCD be a circle, and within it, let two straight-lines, AC and BD, which are not through the center, cut one another at (point) E.
I say that they do not cut one another in half.
For, if possible, let them cut one another in half, such that AE is equal to EC, and BE to ED.
And let the center of the circle ABCD have been found [Prop. 3.1], and let it be (at point) F, and let FE have been joined.
Therefore, since some straight-line through the center, FE, cuts in half some straight-line not through the center, AC, it also cuts it at right-angles [Prop. 3.3].
Thus, FEA is a right-angle.
Again, since some straight-line FE cuts in half some straight-line BD, it also cuts it at right-angles [Prop. 3.3].
Thus, FEB (is) a right-angle.
But FEA was also shown (to be) a right-angle.
Thus, FEA (is) equal to FEB, the lesser to the greater.
The very thing is impossible.
Thus, AC and BD do not cut one another in half.
Thus, in a circle, if two straight-lines, which are not through the center, cut one another then they do not cut one another in half.
(Which is) the very thing it was required to show.
Proposition 5
If two circles cut one another then they will not have the same center.
For let the two circles ABC and CDG cut one another at points B and C.
I say that they will not have the same center.
For, if possible, let E be (the common center), and let EC have been joined, and let EFG have been drawn through (the two circles), at random.
And since point E is the center of the circle ABC, EC is equal to EF.
Again, since point E is the center of the circle CDG, EC is equal to EG.
But EC was also shown (to be) equal to EF.
Thus, EF is also equal to EG, the lesser to the greater.
The very thing is impossible.
Thus, point E is not the (common) center of the circles ABC and CDG.
Thus, if two circles cut one another then they will not have the same center.
(Which is) the very thing it was required to show.
Proposition 6
If two circles touch one another then they will not have the same center.
For let the two circles ABC and CDE touch one another at point C.
I say that they will not have the same center.
For, if possible, let F be (the common center), and let FC have been joined, and let FEB have been drawn through (the two circles), at random.
Therefore, since point F is the center of the circle ABC, FC is equal to FB.
Again, since point F is the center of the circle CDE, FC is equal to FE.
But FC was shown (to be) equal to FB.
Thus, FE is also equal to FB, the lesser to the greater.
The very thing is impossible.
Thus, point F is not the (common) center of the circles ABC and CDE.
Thus, if two circles touch one another then they will not have the same center.
(Which is) the very thing it was required to show.
Proposition 7
If some point, which is not the center of the circle, is taken on the diameter of a circle, and some straight-lines radiate from the point towards the (circumference of the) circle, then the greatest (straight-line) will be that on which the center (lies), and the least the remainder (of the same diameter).
And for the others, a (straight-line) nearer to the (straight-line) through the center is always greater than a (straight-line) further away.
And only two equal (straight-lines) will radiate from the point towards the (circumference of the) circle, (one) on each (side) of the least (straight-line).
Let ABCD be a circle, and let AD be its diameter, and let some point F, which is not the center of the circle, have been taken on AD.
Let E be the center of the circle.
And let some straight-lines, FB, FC, and FG, radiate from F towards (the circumference of) circle ABCD.
I say that FA is the greatest (straight-line), FD the least, and of the others, FB (is) greater than FC, and FC than FG.
For let BE, CE, and GE have been joined.
And since for every triangle (any) two sides are greater than the remaining (side) [Prop. 1.20], EB and EF is thus greater than BF.
And AE (is) equal to BE [thus, BE and EF is equal to AF ].
Thus, AF (is) greater than BF.
Again, since BE is equal to CE, and FE (is) common, the two (straight-lines) BE, EF are equal to the two (straight-lines) CE, EF (respectively).
But, angle BEF (is) also greater than angle CEF.
Thus, the base BF is greater than the base CF.
Thus, the base BF is greater than the base CF [Prop. 1.24].
So, for the same (reasons), CF is also greater than FG.
Again, since GF and FE are greater than EG [Prop. 1.20], and EG (is) equal to ED, GF and FE are thus greater than ED.
Let EF have been taken from both.
Thus, the remainder GF is greater than the remainder FD.
Thus, FA (is) the greatest (straight-line), FD the least, and FB (is) greater than FC, and FC than FG.
I also say that from point F only two equal (straight-lines) will radiate towards (the circumference of) circle ABCD, (one) on each (side) of the least (straight-line) FD.
For let the (angle) FEH, equal to angle GEF, have been constructed on the straight-line EF, at the point E on it [Prop. 1.23], and let FH have been joined.
Therefore, since GE is equal to EH, and EF (is) common, the two (straight-lines) GE, EF are equal to the two (straight-lines) HE, EF (respectively).
And angle GEF (is) equal to angle HEF.
Thus, the base FG is equal to the base FH [Prop. 1.4].
So I say that another (straight-line) equal to FG will not radiate towards (the circumference of) the circle from point F.
For, if possible, let FK (so) radiate.
And since FK is equal to FG, but FH [is equal] to FG, FK is thus also equal to FH, the nearer to the (straight-line) through the center equal to the further away.
The very thing (is) impossible.
Thus, another (straight-line) equal to GF will not radiate from the point F towards (the circumference of) the circle.
Thus, (there is) only one (such straight-line).
Thus, if some point, which is not the center of the circle, is taken on the diameter of a circle, and some straight-lines radiate from the point towards the (circumference of the) circle, then the greatest (straight-line) will be that on which the center (lies), and the least the remainder (of the same diameter).
And for the others, a (straight-line) nearer to the (straight-line) through the center is always greater than a (straight-line) further away.
And only two equal (straight-lines) will radiate from the same point towards the (circumference of the) circle, (one) on each (side) of the least (straight-line).
(Which is) the very thing it was required to show.
Proposition 8
If some point is taken outside a circle, and some straight-lines are drawn from the point to the (circumference of the) circle, one of which (passes) through the center, the remainder (being) random, then for the straight-lines radiating towards the concave (part of the) circumference, the greatest is that (passing) through the center.
For the others, a (straight-line) nearer to the (straight-line) through the center is always greater than one further away.
For the straight-lines radiating towards the convex (part of the) circumference, the least is that between the point and the diameter.
For the others, a (straight-line) nearer to the least (straight-line) is always less than one further away.
And only two equal (straight-lines) will radiate from the point towards the (circumference of the) circle, (one) on each (side) of the least (straight-line).
Let ABC be a circle, and let some point D have been taken outside ABC, and from it let some straight-lines, DA, DE, DF, and DC, have been drawn through (the circle), and let DA be through the center.
I say that for the straight-lines radiating towards the concave (part of the) circumference, AEFC, the greatest is the one (passing) through the center, (namely) AD, and (that) DE (is) greater than DF, and DF than DC.
For the straight-lines radiating towards the convex (part of the) circumference, HLKG, the least is the one between the point and the diameter AG, (namely) DG, and a (straight-line) nearer to the least (straight-line) DG is always less than one farther away, (so that) DK (is less) than DL, and DL than than DH.
For let the center of the circle have been found [Prop. 3.1], and let it be (at point) M [Prop. 3.1].
And let ME, MF, MC, MK, ML, and MH have been joined.
And since AM is equal to EM, let MD have been added to both.
Thus, AD is equal to EM and MD.
But, EM and MD is greater than ED [Prop. 1.20].
Thus, AD is also greater than ED.
Again, since ME is equal to MF, and MD (is) common, the (straight-lines) EM, MD are thus equal to FM, MD.
And angle EMD is greater than angle FMD.
Thus, the base ED is greater than the base FD [Prop. 1.24].
So, similarly, we can show that FD is also greater than CD.
Thus, AD (is) the greatest (straight-line), and DE (is) greater than DF, and DF than DC.
And since MK and KD is greater than MD [Prop. 1.20], and MG (is) equal to MK, the remainder KD is thus greater than the remainder GD.
So GD is less than KD.
And since in triangle MLD, the two internal straight-lines MK and KD were constructed on one of the sides, MD, then MK and KD are thus less than ML and LD [Prop. 1.21].
And MK (is) equal to ML.
Thus, the remainder DK is less than the remainder DL.
So, similarly, we can show that DL is also less than DH.
Thus, DG (is) the least (straight-line), and DK (is) less than DL, and DL than DH.
I also say that only two equal (straight-lines) will radiate from point D towards (the circumference of) the circle, (one) on each (side) on the least (straight-line), DG.
Let the angle DMB, equal to angle KMD, have been constructed on the straight-line MD, at the point M on it [Prop. 1.23], and let DB have been joined.
And since MK is equal to MB, and MD (is) common, the two (straight-lines) KM, MD are equal to the two (straight-lines) BM, MD, respectively.
And angle KMD (is) equal to angle BMD.
Thus, the base DK is equal to the base DB [Prop. 1.4].
[So] I say that another (straight-line) equal to DK will not radiate towards the (circumference of the) circle from point D.
For, if possible, let (such a straight-line) radiate, and let it be DN.
Therefore, since DK is equal to DN, but DK is equal to DB, then DB is thus also equal to DN, (so that) a (straight-line) nearer to the least (straight-line) DG [is] equal to one further away.
The very thing was shown (to be) impossible.
Thus, not more than two equal (straight-lines) will radiate towards (the circumference of) circle ABC from point D, (one) on each side of the least (straight-line) DG.
Thus, if some point is taken outside a circle, and some straight-lines are drawn from the point to the (circumference of the) circle, one of which (passes) through the center, the remainder (being) random, then for the straight-lines radiating towards the concave (part of the) circumference, the greatest is that (passing) through the center.
For the others, a (straight-line) nearer to the (straight-line) through the center is always greater than one further away.
For the straight-lines radiating towards the convex (part of the) circumference, the least is that between the point and the diameter.
For the others, a (straight-line) nearer to the least (straight-line) is always less than one further away.
And only two equal (straight-lines) will radiate from the point towards the (circumference of the) circle, (one) on each (side) of the least (straight-line).
(Which is) the very thing it was required to show.
Proposition 9
If some point is taken inside a circle, and more than two equal straight-lines radiate from the point towards the (circumference of the) circle, then the point taken is the center of the circle.
Let ABC be a circle, and D a point inside it, and let more than two equal straight-lines, DA, DB, and DC, radiate from D towards (the circumference of) circle ABC.
I say that point D is the center of circle ABC.
For let AB and BC have been joined, and (then) have been cut in half at points E and F (respectively) [Prop. 1.10].
And ED and FD being joined, let them have been drawn through to points G, K, H, and L (as EG, DK, FH and DL, respectively).
Therefore, since AE is equal to EB, and ED (is) common, the two (straight-lines) AE, ED are equal to the two (straight-lines) BE, ED (respectively).
And the base DA (is) equal to the base DB.
Thus, angle AED is equal to angle BED [Prop. 1.8].
Thus, angles AED and BED (are) each right-angles [Def. 1.10].
Thus, GK cuts AB in half, and at right-angles.
And since, if some straight-line in a circle cuts some (other) straight-line in half, and at right-angles, then the center of the circle is on the former (straight-line) [Prop. 3.1 corr.], the center of the circle is thus on GK.
So, for the same (reasons), the center of circle ABC is also on HL.
And the straight-lines GK and HL have no common (point) other than point D.
Thus, point D is the center of circle ABC.
Thus, if some point is taken inside a circle, and more than two equal straight-lines radiate from the point towards the (circumference of the) circle, then the point taken is the center of the circle.
(Which is) the very thing it was required to show.
Proposition 10
A circle does not cut a(nother) circle at more than two points.
For, if possible, let the circle ABC cut the circle DEF at more than two points, B, G, F, and H.
And BH and BG being joined, let them (then) have been cut in half at points K and L (respectively).
And KC and LM being drawn at right-angles to BH and BG from K and L (respectively) [Prop. 1.11], let them (then) have been drawn through to points A and E (as CA and ME, respectively).
Therefore, since in circle ABC some straight-line AC cuts some (other) straight-line BH in half, and at right-angles, the center of circle ABC is thus on AC [Prop. 3.1 corr.].
Again, since in the same circle ABC some straight-line NO cuts some (other straight-line) BG in half, and at right-angles, the center of circle ABC is thus on NO [Prop. 3.1 corr.].
And it was also shown (to be) on AC.
And the straight-lines AC and NO meet at no other (point) than P.
Thus, point P is the center of circle ABC.
So, similarly, we can show that P is also the center of circle DEF.
Thus, two circles cutting one another, ABC and DEF, have the same center P.
The very thing is impossible [Prop. 3.5].
Thus, a circle does not cut a(nother) circle at more than two points.
(Which is) the very thing it was required to show.
Proposition 11
If two circles touch one another internally, and their centers are found, then the straight-line joining their centers, being produced, will fall upon the point of union of the circles.
For let two circles, ABC and ADE, touch one another internally at point A, and let the center F of circle ABC have been found [Prop. 3.1], and (the center) G of (circle) ADE [Prop. 3.1].
I say that the straight-line joining G to F, being produced, will fall on A.
For (if) not then, if possible, let it fall like FGH (in the figure), and let AF and AG have been joined.
Therefore, since AG and GF is greater than FA, that is to say FH [Prop. 1.20], let FG have been taken from both.
Thus, the remainder AG is greater than the remainder GH.
And AG (is) equal to GD.
Thus, GD is also greater than GH, the lesser than the greater.
The very thing is impossible.
Thus, the straight-line joining F to G will not fall outside (one circle but inside the other).
Thus, it will fall upon the point of union (of the circles) at point A.
Thus, if two circles touch one another internally, [and their centers are found], then the straight-line joining their centers, [being produced], will fall upon the point of union of the circles.
(Which is) the very thing it was required to show.
Proposition 12
If two circles touch one another externally then the (straight-line) joining their centers will go through the point of union.
For let two circles, ABC and ADE, touch one another externally at point A, and let the center F of ABC have been found [Prop. 3.1], and (the center) G of ADE [Prop. 3.1].
I say that the straight-line joining F to G will go through the point of union at A.
For (if) not then, if possible, let it go like FCDG (in the figure), and let AF and AG have been joined.
Therefore, since point F is the center of circle ABC, FA is equal to FC.
Again, since point G is the center of circle ADE, GA is equal to GD.
And FA was also shown (to be) equal to FC.
Thus, the (straight-lines) FA and AG are equal to the (straight-lines) FC and GD.
So the whole of FG is greater than FA and AG.
But, (it is) also less [Prop. 1.20].
The very thing is impossible.
Thus, the straight-line joining F to G cannot not go through the point of union at A.
Thus, (it will go) through it.
Thus, if two circles touch one another externally then the [straight-line] joining their centers will go through the point of union.
(Which is) the very thing it was required to show.
Proposition 13
A circle does not touch a(nother) circle at more than one point, whether they touch internally or externally.
For, if possible, let circle ABDC touch circle EBFD ---first of all, internally---at more than one point, D and B.
And let the center G of circle ABDC have been found [Prop. 3.1], and (the center) H of EBFD [Prop. 3.1].
Thus, the (straight-line) joining G and H will fall on B and D [Prop. 3.11].
Let it fall like BGHD (in the figure).
And since point G is the center of circle ABDC, BG is equal to GD.
Thus, BG (is) greater than HD.
Thus, BH (is) much greater than HD.
Again, since point H is the center of circle EBFD, BH is equal to HD.
But it was also shown (to be) much greater than it.
The very thing (is) impossible.
Thus, a circle does not touch a(nother) circle internally at more than one point.
So, I say that neither (does it touch) externally (at more than one point).
For, if possible, let circle ACK touch circle ABDC externally at more than one point, A and C.
And let AC have been joined.
Therefore, since two points, A and C, have been taken at random on the circumference of each of the circles ABDC and ACK, the straight-line joining the points will fall inside each (circle) [Prop. 3.2].
But, it fell inside ABDC, and outside ACK [Def. 3.3].
The very thing (is) absurd.
Thus, a circle does not touch a(nother) circle externally at more than one point.
And it was shown that neither (does it) internally.
Thus, a circle does not touch a(nother) circle at more than one point, whether they touch internally or externally.
(Which is) the very thing it was required to show.
Proposition 14
In a circle, equal straight-lines are equally far from the center, and (straight-lines) which are equally far from the center are equal to one another.
Let ABDC be a circle, and let AB and CD be equal straight-lines within it.
I say that AB and CD are equally far from the center.
For let the center of circle ABDC have been found [Prop. 3.1], and let it be (at) E.
And let EF and EG have been drawn from (point) E, perpendicular to AB and CD (respectively) [Prop. 1.12].
And let AE and EC have been joined.
Therefore, since some straight-line, EF, through the center (of the circle), cuts some (other) straight-line, AB, not through the center, at right-angles, it also cuts it in half [Prop. 3.3].
Thus, AF (is) equal to FB.
Thus, AB (is) double AF.
So, for the same (reasons), CD is also double CG.
And AB is equal to CD.
Thus, AF (is) also equal to CG.
And since AE is equal to EC, the (square) on AE (is) also equal to the (square) on EC.
But, the (sum of the squares) on AF and EF (is) equal to the (square) on AE.
For the angle at F (is) a right-angle [Prop. 1.47].
And the (sum of the squares) on EG and GC (is) equal to the (square) on EC.
For the angle at G (is) a right-angle [Prop. 1.47].
Thus, the (sum of the squares) on AF and FE is equal to the (sum of the squares) on CG and GE, of which the (square) on AF is equal to the (square) on CG.
For AF is equal to CG.
Thus, the remaining (square) on FE is equal to the (remaining square) on EG.
Thus, EF (is) equal to EG.
And straight-lines in a circle are said to be equally far from the center when perpendicular (straight-lines) which are drawn to them from the center are equal [Def. 3.4].
Thus, AB and CD are equally far from the center.
So, let the straight-lines AB and CD be equally far from the center.
That is to say, let EF be equal to EG.
I say that AB is also equal to CD.
For, with the same construction, we can, similarly, show that AB is double AF, and CD (double) CG.
And since AE is equal to CE, the (square) on AE is equal to the (square) on CE.
But, the (sum of the squares) on EF and FA is equal to the (square) on AE [Prop. 1.47].
And the (sum of the squares) on EG and GC (is) equal to the (square) on CE [Prop. 1.47].
Thus, the (sum of the squares) on EF and FA is equal to the (sum of the squares) on EG and GC, of which the (square) on EF is equal to the (square) on EG.
For EF (is) equal to EG.
Thus, the remaining (square) on AF is equal to the (remaining square) on CG.
Thus, AF (is) equal to CG.
And AB is double AF, and CD double CG.
Thus, AB (is) equal to CD.
Thus, in a circle, equal straight-lines are equally far from the center, and (straight-lines) which are equally far from the center are equal to one another.
(Which is) the very thing it was required to show.
Proposition 15
In a circle, a diameter (is) the greatest (straight-line), and for the others, a (straight-line) nearer to the center is always greater than one further away.
Let ABCD be a circle, and let AD be its diameter, and E (its) center.
And let BC be nearer to the diameter AD, and FG further away.
I say that AD is the greatest (straight-line), and BC (is) greater than FG.
For let EH and EK have been drawn from the center E, at right-angles to BC and FG (respectively) [Prop. 1.12].
And since BC is nearer to the center, and FG further away, EK (is) thus greater than EH [Def. 3.5].
Let EL be made equal to EH [Prop. 1.3].
And LM being drawn through L, at right-angles to EK [Prop. 1.11], let it have been drawn through to N (as MN).
And let ME, EN, FE, and EG have been joined.
And since EH is equal to EL, BC is also equal to MN [Prop. 3.14].
Again, since AE is equal to EM, and ED to EN, AD is thus equal to ME and EN.
But, ME and EN is greater than MN [Prop. 1.20], and MN (is) equal to BC.
Thus, AD is greater than BC.
And since the two (straight-lines) ME, EN are equal to the two (straight-lines) FE, EG (respectively), and angle MEN [is] greater than angle FEG, the base MN is thus greater than the base FG [Prop. 1.24].
But, MN was shown (to be) equal to BC.
Thus, the diameter AD (is) the greatest (straight-line), and BC (is) greater than FG.
Thus, in a circle, a diameter (is) the greatest (straight-line), and for the others, a (straight-line) nearer to the center is always greater than one further away.
(Which is) the very thing it was required to show.
Proposition 16
A (straight-line) drawn at right-angles to the diameter of a circle, from its end, will fall outside the circle.
And another straight-line cannot be inserted into the space between the (aforementioned) straight-line and the circumference.
And the angle of the semi-circle is greater than any acute rectilinear angle whatsoever, and the remaining (angle is) less (than any acute rectilinear angle).
Let ABC be a circle around the center D and the diameter AB.
I say that the (straight-line) drawn from A, at right-angles to AB [Prop. 1.11], from its end, will fall outside the circle.
For (if) not then, if possible, let it fall inside, like CA (in the figure), and let DC have been joined.
Since DA is equal to DC, angle DAC is also equal to angle ACD [Prop. 1.5].
And DAC (is) a right-angle.
Thus, ACD (is) also a right-angle.
So, in triangle ACD, the two angles DAC and ACD are equal to two right-angles.
The very thing is impossible [Prop. 1.17].
Thus, the (straight-line) drawn from point A, at right-angles to BA, will not fall inside the circle.
So, similarly, we can show that neither (will it fall) on the circumference.
Thus, (it will fall) outside (the circle).
Let it fall like AE (in the figure).
So, I say that another straight-line cannot be inserted into the space between the straight-line AE and the circumference CHA.
For, if possible, let it be inserted like FA (in the figure), and let DG have been drawn from point D, perpendicular to FA [Prop. 1.12].
And since AGD is a right-angle, and DAG (is) less than a right-angle, AD (is) thus greater than DG [Prop. 1.19].
And DA (is) equal to DH.
Thus, DH (is) greater than DG, the lesser than the greater.
The very thing is impossible.
Thus, another straight-line cannot be inserted into the space between the straight-line ( AE) and the circumference.
And I also say that the semi-circular angle contained by the straight-line BA and the circumference CHA is greater than any acute rectilinear angle whatsoever, and the remaining (angle) contained by the circumference CHA and the straight-line AE is less than any acute rectilinear angle whatsoever.
For if any rectilinear angle is greater than the (angle) contained by the straight-line BA and the circumference CHA, or less than the (angle) contained by the circumference CHA and the straight-line AE, then a straight-line can be inserted into the space between the circumference CHA and the straight-line AE ---anything which will make (an angle) contained by straight-lines greater than the angle contained by the straight-line BA and the circumference CHA, or less than the (angle) contained by the circumference CHA and the straight-line AE.
But (such a straight-line) cannot be inserted.
Thus, an acute (angle) contained by straight-lines cannot be greater than the angle contained by the straight-line BA and the circumference CHA, neither (can it be) less than the (angle) contained by the circumference CHA and the straight-line AE.
Corollary
So, from this, (it is) manifest that a (straight-line) drawn at right-angles to the diameter of a circle, from its extremity, touches the circle [and that the straight-line touches the circle at a single point, inasmuch as it was also shown that a (straight-line) meeting (the circle) at two (points) falls inside it [Prop. 3.2] ].
(Which is) the very thing it was required to show.
Proposition 17
To draw a straight-line touching a given circle from a given point.
Let A be the given point, and BCD the given circle.
So it is required to draw a straight-line touching circle BCD from point A.
For let the center E of the circle have been found [Prop. 3.1], and let AE have been joined.
And let (the circle) AFG have been drawn with center E and radius EA.
And let DF have been drawn from from (point) D, at right-angles to EA [Prop. 1.11].
And let EF and AB have been joined.
I say that the (straight-line) AB has been drawn from point A touching circle BCD.
For since E is the center of circles BCD and AFG, EA is thus equal to EF, and ED to EB.
So the two (straight-lines) AE, EB are equal to the two (straight-lines) FE, ED (respectively).
And they contain a common angle at E.
Thus, the base DF is equal to the base AB, and triangle DEF is equal to triangle EBA, and the remaining angles (are equal) to the (corresponding) remaining angles [Prop. 1.4].
Thus, (angle) EDF (is) equal to EBA.
And EDF (is) a right-angle.
Thus, EBA (is) also a right-angle.
And EB is a radius.
And a (straight-line) drawn at right-angles to the diameter of a circle, from its extremity, touches the circle [Prop. 3.16 corr.].
Thus, AB touches circle BCD.
Thus, the straight-line AB has been drawn touching the given circle BCD from the given point A.
(Which is) the very thing it was required to do.
Proposition 18
If some straight-line touches a circle, and some (other) straight-line is joined from the center (of the circle) to the point of contact, then the (straight-line) so joined will be perpendicular to the tangent.
For let some straight-line DE touch the circle ABC at point C, and let the center F of circle ABC have been found [Prop. 3.1], and let FC have been joined from F to C.
I say that FC is perpendicular to DE.
For if not, let FG have been drawn from F, perpendicular to DE [Prop. 1.12].
Therefore, since angle FGC is a right-angle, (angle) FCG is thus acute [Prop. 1.17].
And the greater angle is subtended by the greater side [Prop. 1.19].
Thus, FC (is) greater than FG.
And FC (is) equal to FB.
Thus, FB (is) also greater than FG, the lesser than the greater.
The very thing is impossible.
Thus, FG is not perpendicular to DE.
So, similarly, we can show that neither (is) any other (straight-line) except FC.
Thus, FC is perpendicular to DE.
Thus, if some straight-line touches a circle, and some (other) straight-line is joined from the center (of the circle) to the point of contact, then the (straight-line) so joined will be perpendicular to the tangent.
(Which is) the very thing it was required to show.
Proposition 19
If some straight-line touches a circle, and a straight-line is drawn from the point of contact, at right-[angles] to the tangent, then the center (of the circle) will be on the (straight-line) so drawn.
For let some straight-line DE touch the circle ABC at point C.
And let CA have been drawn from C, at right-angles to DE [Prop. 1.11].
I say that the center of the circle is on AC.
For (if) not, if possible, let F be (the center of the circle), and let CF have been joined.
[Therefore], since some straight-line DE touches the circle ABC, and FC has been joined from the center to the point of contact, FC is thus perpendicular to DE [Prop. 3.18].
Thus, FCE is a right-angle.
And ACE is also a right-angle.
Thus, FCE is equal to ACE, the lesser to the greater.
The very thing is impossible.
Thus, F is not the center of circle ABC.
So, similarly, we can show that neither is any (point) other (than one) on AC.
Thus, if some straight-line touches a circle, and a straight-line is drawn from the point of contact, at right-angles to the tangent, then the center (of the circle) will be on the (straight-line) so drawn.
(Which is) the very thing it was required to show.
Proposition 20
In a circle, the angle at the center is double that at the circumference, when the angles have the same circumference base.
Let ABC be a circle, and let BEC be an angle at its center, and BAC (one) at (its) circumference.
And let them have the same circumference base BC.
I say that angle BEC is double (angle) BAC.
For being joined, let AE have been drawn through to F (as AF).
Therefore, since EA is equal to EB, angle EAB (is) also equal to EBA [Prop. 1.5].
Thus, angle EAB and EBA is double (angle) EAB.
And BEF (is) equal to EAB and EBA [Prop. 1.32].
Thus, BEF is also double EAB.
So, for the same (reasons), FEC is also double EAC.
Thus, the whole (angle) BEC is double the whole (angle) BAC.
So let another (straight-line) have been inflected, and let there be another angle, BDC.
And DE being joined, let it have been produced to G.
So, similarly, we can show that angle GEC is double EDC, of which GEB is double EDB.
Thus, the remaining (angle) BEC is double the (remaining angle) BDC.
Thus, in a circle, the angle at the center is double that at the circumference, when [the angles] have the same circumference base.
(Which is) the very thing it was required to show.
Proposition 21
In a circle, angles in the same segment are equal to one another.
Let ABCD be a circle, and let BAD and BED be angles in the same segment BAED.
I say that angles BAD and BED are equal to one another.
For let the center of circle ABCD have been found [Prop. 3.1], and let it be (at point) F.
And let BF and FD have been joined.
And since angle BFD is at the center, and BAD at the circumference, and they have the same circumference base BCD, angle BFD is thus double BAD [Prop. 3.20].
So, for the same (reasons), BFD is also double BED.
Thus, BAD (is) equal to BED.
Thus, in a circle, angles in the same segment are equal to one another.
(Which is) the very thing it was required to show.
Proposition 22
For quadrilaterals within circles, the (sum of the) opposite angles is equal to two right-angles.
Let ABCD be a circle, and let ABCD be a quadrilateral within it.
I say that the (sum of the) opposite angles is equal to two right-angles.
Let AC and BD have been joined.
Therefore, since the three angles of any triangle are equal to two right-angles [Prop. 1.32], the three angles CAB, ABC, and BCA of triangle ABC are thus equal to two right-angles.
And CAB (is) equal to BDC.
For they are in the same segment BADC [Prop. 3.21].
And ACB (is equal) to ADB.
For they are in the same segment ADCB [Prop. 3.21].
Thus, the whole of ADC is equal to BAC and ACB.
Let ABC have been added to both.
Thus, ABC, BAC, and ACB are equal to ABC and ADC.
But, ABC, BAC, and ACB are equal to two right-angles.
Thus, ABC and ADC are also equal to two right-angles.
Similarly, we can show that angles BAD and DCB are also equal to two right-angles.
Thus, for quadrilaterals within circles, the (sum of the) opposite angles is equal to two right-angles.
(Which is) the very thing it was required to show.
Proposition 23
Two similar and unequal segments of circles cannot be constructed on the same side of the same straight-line.
For, if possible, let the two similar and unequal segments of circles, ACB and ADB, have been constructed on the same side of the same straight-line AB.
And let ACD have been drawn through (the segments), and let CB and DB have been joined.
Therefore, since segment ACB is similar to segment ADB, and similar segments of circles are those accepting equal angles [Def. 3.11], angle ACB is thus equal to ADB, the external to the internal.
The very thing is impossible [Prop. 1.16].
Thus, two similar and unequal segments of circles cannot be constructed on the same side of the same straight-line.
Proposition 24
Similar segments of circles on equal straight-lines are equal to one another.
For let AEB and CFD be similar segments of circles on the equal straight-lines AB and CD (respectively).
I say that segment AEB is equal to segment CFD.
For if the segment AEB is applied to the segment CFD, and point A is placed on (point) C, and the straight-line AB on CD, then point B will also coincide with point D, on account of AB being equal to CD.
And if AB coincides with CD then the segment AEB will also coincide with CFD.
For if the straight-line AB coincides with CD, and the segment AEB does not coincide with CFD, then it will surely either fall inside it, outside (it), or it will miss like CGD (in the figure), and a circle (will) cut (another) circle at more than two points.
The very thing is impossible [Prop. 3.10].
Thus, if the straight-line AB is applied to CD, the segment AEB cannot not also coincide with CFD.
Thus, it will coincide, and will be equal to it [C.N. 4].
Thus, similar segments of circles on equal straight-lines are equal to one another.
(Which is) the very thing it was required to show.
Proposition 25
For a given segment of a circle, to complete the circle, the very one of which it is a segment.
Let ABC be the given segment of a circle.
So it is required to complete the circle for segment ABC, the very one of which it is a segment.
For let AC have been cut in half at (point) D [Prop. 1.10], and let DB have been drawn from point D, at right-angles to AC [Prop. 1.11].
And let AB have been joined.
Thus, angle ABD is surely either greater than, equal to, or less than (angle) BAD.
First of all, let it be greater.
And let (angle) BAE, equal to angle ABD, have been constructed on the straight-line BA, at the point A on it [Prop. 1.23].
And let DB have been drawn through to E (as BE), and let EC have been joined.
Therefore, since angle ABE is equal to BAE, the straight-line EB is thus also equal to EA [Prop. 1.6].
And since AD is equal to DC, and DE (is) common, the two (straight-lines) AD, DE are equal to the two (straight-lines) CD, DE, respectively.
And angle ADE is equal to angle CDE.
For each (is) a right-angle.
Thus, the base AE is equal to the base CE [Prop. 1.4].
But, AE was shown (to be) equal to BE.
Thus, BE is also equal to CE.
Thus, the three (straight-lines) AE, EB, and EC are equal to one another.
Thus, if a circle is drawn with center E, and radius one of AE, EB, or EC, it will also go through the remaining points (of the segment), and the (associated circle) will have been completed [Prop. 3.9].
Thus, a circle has been completed from the given segment of a circle.
And (it is) clear that the segment ABC is less than a semi-circle, because the center E happens to lie outside it.
[And], similarly, even if angle ABD is equal to BAD, (since) AD becomes equal to each of BD [Prop. 1.6] and DC, the three (straight-lines) DA, DB, and DC will be equal to one another.
And point D will be the center of the completed circle.
And ABC will manifestly be a semi-circle.
And if ABD is less than BAD, and we construct (angle BAE), equal to angle ABD, on the straight-line BA, at the point A on it [Prop. 1.23], then the center will fall on DB, inside the segment ABC.
And segment ABC will manifestly be greater than a semi-circle.
Thus, a circle has been completed from the given segment of a circle.
(Which is) the very thing it was required to do.
Proposition 26
In equal circles, equal angles stand upon equal circumferences whether they are standing at the center or at the circumference.
Let ABC and DEF be equal circles, and within them let BGC and EHF be equal angles at the center, and BAC and EDF (equal angles) at the circumference.
I say that circumference BKC is equal to circumference ELF.
For let BC and EF have been joined.
And since circles ABC and DEF are equal, their radii are equal.
So the two (straight-lines) BG, GC (are) equal to the two (straight-lines) EH, HF (respectively).
And the angle at G (is) equal to the angle at H.
Thus, the base BC is equal to the base EF [Prop. 1.4].
And since the angle at A is equal to the (angle) at D, the segment BAC is thus similar to the segment EDF [Def. 3.11].
And they are on equal straight-lines [ BC and EF ].
And similar segments of circles on equal straight-lines are equal to one another [Prop. 3.24].
Thus, segment BAC is equal to (segment) EDF.
And the whole circle ABC is also equal to the whole circle DEF.
Thus, the remaining circumference BKC is equal to the (remaining) circumference ELF.
Thus, in equal circles, equal angles stand upon equal circumferences, whether they are standing at the center or at the circumference.
(Which is) the very thing which it was required to show.
Proposition 27
In equal circles, angles standing upon equal circumferences are equal to one another, whether they are standing at the center or at the circumference.
For let the angles BGC and EHF at the centers G and H, and the (angles) BAC and EDF at the circumferences, stand upon the equal circumferences BC and EF, in the equal circles ABC and DEF (respectively).
I say that angle BGC is equal to (angle) EHF, and BAC is equal to EDF.
For if BGC is unequal to EHF, one of them is greater.
Let BGC be greater, and let the (angle) BGK, equal to angle EHF, have been constructed on the straight-line BG, at the point G on it [Prop. 1.23].
But equal angles (in equal circles) stand upon equal circumferences, when they are at the centers [Prop. 3.26].
Thus, circumference BK (is) equal to circumference EF.
But, EF is equal to BC.
Thus, BK is also equal to BC, the lesser to the greater.
The very thing is impossible.
Thus, angle BGC is not unequal to EHF.
Thus, (it is) equal.
And the (angle) at A is half BGC, and the (angle) at D half EHF [Prop. 3.20].
Thus, the angle at A (is) also equal to the (angle) at D.
Thus, in equal circles, angles standing upon equal circumferences are equal to one another, whether they are standing at the center or at the circumference.
(Which is) the very thing it was required to show.
Proposition 28
In equal circles, equal straight-lines cut off equal circumferences, the greater (circumference being equal) to the greater, and the lesser to the lesser.
Let ABC and DEF be equal circles, and let AB and DE be equal straight-lines in these circles, cutting off the greater circumferences ACB and DFE, and the lesser (circumferences) AGB and DHE (respectively).
I say that the greater circumference ACB is equal to the greater circumference DFE, and the lesser circumference AGB to (the lesser) DHE.
For let the centers of the circles, K and L, have been found [Prop. 3.1], and let AK, KB, DL, and LE have been joined.
And since ( ABC and DEF) are equal circles, their radii are also equal [Def. 3.1].
So the two (straight-lines) AK, KB are equal to the two (straight-lines) DL, LE (respectively).
And the base AB (is) equal to the base DE.
Thus, angle AKB is equal to angle DLE [Prop. 1.8].
And equal angles stand upon equal circum-ferences, when they are at the centers [Prop. 3.26].
Thus, circumference AGB (is) equal to DHE.
And the whole circle ABC is also equal to the whole circle DEF.
Thus, the remaining circumference ACB is also equal to the remaining circumference DFE.
Thus, in equal circles, equal straight-lines cut off equal circumferences, the greater (circumference being equal) to the greater, and the lesser to the lesser.
(Which is) the very thing it was required to show.
Proposition 29
In equal circles, equal straight-lines subtend equal circumferences.
Let ABC and DEF be equal circles, and within them let the equal circumferences BGC and EHF have been cut off.
And let the straight-lines BC and EF have been joined.
I say that BC is equal to EF.
For let the centers of the circles have been found [Prop. 3.1], and let them be (at) K and L.
And let BK, KC, EL, and LF have been joined.
And since the circumference BGC is equal to the circumference EHF, the angle BKC is also equal to (angle) ELF [Prop. 3.27].
And since the circles ABC and DEF are equal, their radii are also equal [Def. 3.1].
So the two (straight-lines) BK, KC are equal to the two (straight-lines) EL, LF (respectively).
And they contain equal angles.
Thus, the base BC is equal to the base EF [Prop. 1.4].
Thus, in equal circles, equal straight-lines subtend equal circumferences.
(Which is) the very thing it was required to show.
Proposition 30
To cut a given circumference in half.
Let ADB be the given circumference.
So it is required to cut circumference ADB in half.
Let AB have been joined, and let it have been cut in half at (point) C [Prop. 1.10].
And let CD have been drawn from point C, at right-angles to AB [Prop. 1.11].
And let AD, and DB have been joined.
And since AC is equal to CB, and CD (is) common, the two (straight-lines) AC, CD are equal to the two (straight-lines) BC, CD (respectively).
And angle ACD (is) equal to angle BCD.
For (they are) each right-angles.
Thus, the base AD is equal to the base DB [Prop. 1.4].
And equal straight-lines cut off equal circumferences, the greater (circumference being equal) to the greater, and the lesser to the lesser [Prop. 1.28].
And the circumferences AD and DB are each less than a semi-circle.
Thus, circumference AD (is) equal to circumference DB.
Thus, the given circumference has been cut in half at point D.
(Which is) the very thing it was required to do.
Proposition 31
In a circle, the angle in a semi-circle is a right-angle, and that in a greater segment (is) less than a right-angle, and that in a lesser segment (is) greater than a right-angle.
And, further, the angle of a segment greater (than a semi-circle) is greater than a right-angle, and the angle of a segment less (than a semi-circle) is less than a right-angle.
Let ABCD be a circle, and let BC be its diameter, and E its center.
And let BA, AC, AD, and DC have been joined.
I say that the angle BAC in the semi-circle BAC is a right-angle, and the angle ABC in the segment ABC, (which is) greater than a semi-circle, is less than a right-angle, and the angle ADC in the segment ADC, (which is) less than a semi-circle, is greater than a right-angle.
Let AE have been joined, and let BA have been drawn through to F (as BF).
And since BE is equal to EA, angle ABE is also equal to BAE [Prop. 1.5].
Again, since CE is equal to EA, ACE is also equal to CAE [Prop. 1.5].
Thus, the whole (angle) BAC is equal to the two (angles) ABC and ACB.
And FAC, (which is) external to triangle ABC, is also equal to the two angles ABC and ACB [Prop. 1.32].
Thus, angle BAC (is) also equal to FAC.
Thus, (they are) each right-angles [Def. 1.10].
Thus, the angle BAC in the semi-circle BAC is a right-angle.
And since the two angles ABC and BAC of triangle ABC are less than two right-angles [Prop. 1.17], and BAC is a right-angle, angle ABC is thus less than a right-angle.
And it is in segment ABC, (which is) greater than a semi-circle.
And since ABCD is a quadrilateral within a circle, and for quadrilaterals within circles the (sum of the) opposite angles is equal to two right-angles [Prop. 3.22] [angles ABC and ADC are thus equal to two right-angles], and (angle) ABC is less than a right-angle.
The remaining angle ADC is thus greater than a right-angle.
And it is in segment ADC, (which is) less than a semi-circle.
I also say that the angle of the greater segment, (namely) that contained by the circumference ABC and the straight-line AC, is greater than a right-angle.
And the angle of the lesser segment, (namely) that contained by the circumference ADC and the straight-line AC, is less than a right-angle.
And this is immediately apparent.
For since the (angle contained by) the two straight-lines BA and AC is a right-angle, the (angle) contained by the circumference ABC and the straight-line AC is thus greater than a right-angle.
Again, since the (angle contained by) the straight-lines AC and AF is a right-angle, the (angle) contained by the circumference ADC and the straight-line CA is thus less than a right-angle.
Thus, in a circle, the angle in a semi-circle is a right-angle, and that in a greater segment (is) less than a right-angle, and that in a lesser [segment] (is) greater than a right-angle.
And, further, the [angle] of a segment greater (than a semi-circle) [is] greater than a right-angle, and the [angle] of a segment less (than a semi-circle) is less than a right-angle.
(Which is) the very thing it was required to show.
Proposition 32
If some straight-line touches a circle, and some (other) straight-line is drawn across, from the point of contact into the circle, cutting the circle (in two), then those angles the (straight-line) makes with the tangent will be equal to the angles in the alternate segments of the circle.
For let some straight-line EF touch the circle ABCD at the point B, and let some (other) straight-line BD have been drawn from point B into the circle ABCD, cutting it (in two).
I say that the angles BD makes with the tangent EF will be equal to the angles in the alternate segments of the circle.
That is to say, that angle FBD is equal to the angle constructed in segment BAD, and angle EBD is equal to the angle constructed in segment DCB.
For let BA have been drawn from B, at right-angles to EF [Prop. 1.11].
And let the point C have been taken at random on the circumference BD.
And let AD, DC, and CB have been joined.
And since some straight-line EF touches the circle ABCD at point B, and BA has been drawn from the point of contact, at right-angles to the tangent, the center of circle ABCD is thus on BA [Prop. 3.19].
Thus, BA is a diameter of circle ABCD.
Thus, angle ADB, being in a semi-circle, is a right-angle [Prop. 3.31].
Thus, the remaining angles (of triangle ADB) BAD and ABD are equal to one right-angle [Prop. 1.32].
And ABF is also a right-angle.
Thus, ABF is equal to BAD and ABD.
Let ABD have been subtracted from both.
Thus, the remaining angle DBF is equal to the angle BAD in the alternate segment of the circle.
And since ABCD is a quadrilateral in a circle, (the sum of) its opposite angles is equal to two right-angles [Prop. 3.22].
And DBF and DBE is also equal to two right-angles [Prop. 1.13].
Thus, DBF and DBE is equal to BAD and BCD, of which BAD was shown (to be) equal to DBF.
Thus, the remaining (angle) DBE is equal to the angle DCB in the alternate segment DCB of the circle.
Thus, if some straight-line touches a circle, and some (other) straight-line is drawn across, from the point of contact into the circle, cutting the circle (in two), then those angles the (straight-line) makes with the tangent will be equal to the angles in the alternate segments of the circle.
(Which is) the very thing it was required to show.
Proposition 33
To draw a segment of a circle, accepting an angle equal to a given rectilinear angle, on a given straight-line.
Let AB be the given straight-line, and C the given rectilinear angle.
So it is required to draw a segment of a circle, accepting an angle equal to C, on the given straight-line AB.
So the [angle] C is surely either acute, a right-angle, or obtuse.
First of all, let it be acute.
And, as in the first diagram (from the left), let (angle) BAD, equal to angle C, have been constructed on the straight-line AB, at the point A (on it) [Prop. 1.23].
Thus, BAD is also acute.
Let AE have been drawn, at right-angles to DA [Prop. 1.11].
And let AB have been cut in half at F [Prop. 1.10].
And let FG have been drawn from point F, at right-angles to AB [Prop. 1.11].
And let GB have been joined.
And since AF is equal to FB, and FG (is) common, the two (straight-lines) AF, FG are equal to the two (straight-lines) BF, FG (respectively).
And angle AFG (is) equal to [angle] BFG.
Thus, the base AG is equal to the base BG [Prop. 1.4].
Thus, the circle drawn with center G, and radius GA, will also go through B (as well as A).
Let it have been drawn, and let it be (denoted) ABE.
And let EB have been joined.
Therefore, since AD is at the extremity of diameter AE, (namely, point) A, at right-angles to AE, the (straight-line) AD thus touches the circle ABE [Prop. 3.16 corr.].
Therefore, since some straight-line AD touches the circle ABE, and some (other) straight-line AB has been drawn across from the point of contact A into circle ABE, angle DAB is thus equal to the angle AEB in the alternate segment of the circle [Prop. 3.32].
But, DAB is equal to C.
Thus, angle C is also equal to AEB.
Thus, a segment AEB of a circle, accepting the angle AEB (which is) equal to the given (angle) C, has been drawn on the given straight-line AB.
And so let C be a right-angle.
And let it again be necessary to draw a segment of a circle on AB, accepting an angle equal to the right-[angle] C.
Let the (angle) BAD [again] have been constructed, equal to the right-angle C [Prop. 1.23], as in the second diagram (from the left).
And let AB have been cut in half at F [Prop. 1.10].
And let the circle AEB have been drawn with center F, and radius either FA or FB.
Thus, the straight-line AD touches the circle ABE, on account of the angle at A being a right-angle [Prop. 3.16 corr.].
And angle BAD is equal to the angle in segment AEB.
For (the latter angle), being in a semi-circle, is also a right-angle [Prop. 3.31].
But, BAD is also equal to C.
Thus, the (angle) in (segment) AEB is also equal to C.
Thus, a segment AEB of a circle, accepting an angle equal to C, has again been drawn on AB.
And so let (angle) C be obtuse.
And let (angle) BAD, equal to ( C), have been constructed on the straight-line AB, at the point A (on it) [Prop. 1.23], as in the third diagram (from the left).
And let AE have been drawn, at right-angles to AD [Prop. 1.11].
And let AB have again been cut in half at F [Prop. 1.10].
And let FG have been drawn, at right-angles to AB [Prop. 1.10].
And let GB have been joined.
And again, since AF is equal to FB, and FG (is) common, the two (straight-lines) AF, FG are equal to the two (straight-lines) BF, FG (respectively).
And angle AFG (is) equal to angle BFG.
Thus, the base AG is equal to the base BG [Prop. 1.4].
Thus, a circle of center G, and radius GA, being drawn, will also go through B (as well as A).
Let it go like AEB (in the third diagram from the left).
And since AD is at right-angles to the diameter AE, at its extremity, AD thus touches circle AEB [Prop. 3.16 corr.].
And AB has been drawn across (the circle) from the point of contact A.
Thus, angle BAD is equal to the angle constructed in the alternate segment AHB of the circle [Prop. 3.32].
But, angle BAD is equal to C.
Thus, the angle in segment AHB is also equal to C.
Thus, a segment AHB of a circle, accepting an angle equal to C, has been drawn on the given straight-line AB.
(Which is) the very thing it was required to do.
Proposition 34
To cut off a segment, accepting an angle equal to a given rectilinear angle, from a given circle.
Let ABC be the given circle, and D the given rectilinear angle.
So it is required to cut off a segment, accepting an angle equal to the given rectilinear angle D, from the given circle ABC.
Let EF have been drawn touching ABC at point B.
And let (angle) FBC, equal to angle D, have been constructed on the straight-line FB, at the point B on it [Prop. 1.23].
Therefore, since some straight-line EF touches the circle ABC, and BC has been drawn across (the circle) from the point of contact B, angle FBC is thus equal to the angle constructed in the alternate segment BAC [Prop. 1.32].
But, FBC is equal to D.
Thus, the (angle) in the segment BAC is also equal to [angle] D.
Thus, the segment BAC, accepting an angle equal to the given rectilinear angle D, has been cut off from the given circle ABC.
(Which is) the very thing it was required to do.
Proposition 35
If two straight-lines in a circle cut one another then the rectangle contained by the pieces of one is equal to the rectangle contained by the pieces of the other.
For let the two straight-lines AC and BD, in the circle ABCD, cut one another at point E.
I say that the rectangle contained by AE and EC is equal to the rectangle contained by DE and EB.
In fact, if AC and BD are through the center (as in the first diagram from the left), so that E is the center of circle ABCD, then (it is) clear that, AE, EC, DE, and EB being equal, the rectangle contained by AE and EC is also equal to the rectangle contained by DE and EB.
So let AC and DB not be though the center (as in the second diagram from the left), and let the center of ABCD have been found [Prop. 3.1], and let it be (at) F.
And let FG and FH have been drawn from F, perpendicular to the straight-lines AC and DB (respectively) [Prop. 1.12].
And let FB, FC, and FE have been joined.
And since some straight-line, GF, through the center, cuts at right-angles some (other) straight-line, AC, not through the center, then it also cuts it in half [Prop. 3.3].
Thus, AG (is) equal to GC.
Therefore, since the straight-line AC is cut equally at G, and unequally at E, the rectangle contained by AE and EC plus the square on EG is thus equal to the (square) on GC [Prop. 2.5].
Let the (square) on GF have been added [to both].
Thus, the (rectangle contained) by AE and EC plus the (sum of the squares) on GE and GF is equal to the (sum of the squares) on CG and GF.
But, the (square) on FE is equal to the (sum of the squares) on EG and GF [Prop. 1.47], and the (square) on FC is equal to the (sum of the squares) on CG and GF [Prop. 1.47].
Thus, the (rectangle contained) by AE and EC plus the (square) on FE is equal to the (square) on FC.
And FC (is) equal to FB.
Thus, the (rectangle contained) by AE and EC plus the (square) on FE is equal to the (square) on FB.
So, for the same (reasons), the (rectangle contained) by DE and EB plus the (square) on FE is equal to the (square) on FB.
And the (rectangle contained) by AE and EC plus the (square) on FE was also shown (to be) equal to the (square) on FB.
Thus, the (rectangle contained) by AE and EC plus the (square) on FE is equal to the (rectangle contained) by DE and EB plus the (square) on FE.
Let the (square) on FE have been taken from both.
Thus, the remaining rectangle contained by AE and EC is equal to the rectangle contained by DE and EB.
Thus, if two straight-lines in a circle cut one another then the rectangle contained by the pieces of one is equal to the rectangle contained by the pieces of the other.
(Which is) the very thing it was required to show.
Proposition 36
If some point is taken outside a circle, and two straight-lines radiate from it towards the circle, and (one) of them cuts the circle, and the (other) touches (it), then the (rectangle contained) by the whole (straight-line) cutting (the circle), and the (part of it) cut off outside (the circle), between the point and the convex circumference, will be equal to the square on the tangent (line).
For let some point D have been taken outside circle ABC, and let two straight-lines, DCA and DB, radiate from D towards circle ABC.
And let DCA cut circle ABC, and let BD touch (it).
I say that the rectangle contained by AD and DC is equal to the square on DB.
DCA is surely either through the center, or not.
Let it first of all be through the center, and let F be the center of circle ABC, and let FB have been joined.
Thus, (angle) FBD is a right-angle [Prop. 3.18].
And since straight-line AC is cut in half at F, let CD have been added to it.
Thus, the (rectangle contained) by AD and DC plus the (square) on FC is equal to the (square) on FD [Prop. 2.6].
And FC (is) equal to FB.
Thus, the (rectangle contained) by AD and DC plus the (square) on FB is equal to the (square) on FD.
And the (square) on FD is equal to the (sum of the squares) on FB and BD [Prop. 1.47].
Thus, the (rectangle contained) by AD and DC plus the (square) on FB is equal to the (sum of the squares) on FB and BD.
Let the (square) on FB have been subtracted from both.
Thus, the remaining (rectangle contained) by AD and DC is equal to the (square) on the tangent DB.
And so let DCA not be through the center of circle ABC, and let the center E have been found, and let EF have been drawn from E, perpendicular to AC [Prop. 1.12].
And let EB, EC, and ED have been joined.
(Angle) EBD (is) thus a right-angle [Prop. 3.18].
And since some straight-line, EF, through the center, cuts some (other) straight-line, AC, not through the center, at right-angles, it also cuts it in half [Prop. 3.3].
Thus, AF is equal to FC.
And since the straight-line AC is cut in half at point F, let CD have been added to it.
Thus, the (rectangle contained) by AD and DC plus the (square) on FC is equal to the (square) on FD [Prop. 2.6].
Let the (square) on FE have been added to both.
Thus, the (rectangle contained) by AD and DC plus the (sum of the squares) on CF and FE is equal to the (sum of the squares) on FD and FE.
But the (square) on EC is equal to the (sum of the squares) on CF and FE.
For [angle] EFC [is] a right-angle [Prop. 1.47].
And the (square) on ED is equal to the (sum of the squares) on DF and FE [Prop. 1.47].
Thus, the (rectangle contained) by AD and DC plus the (square) on EC is equal to the (square) on ED.
And EC (is) equal to EB.
Thus, the (rectangle contained) by AD and DC plus the (square) on EB is equal to the (square) on ED.
And the (sum of the squares) on EB and BD is equal to the (square) on ED.
For EBD (is) a right-angle [Prop. 1.47].
Thus, the (rectangle contained) by AD and DC plus the (square) on EB is equal to the (sum of the squares) on EB and BD.
Let the (square) on EB have been subtracted from both.
Thus, the remaining (rectangle contained) by AD and DC is equal to the (square) on BD.
Thus, if some point is taken outside a circle, and two straight-lines radiate from it towards the circle, and (one) of them cuts the circle, and (the other) touches (it), then the (rectangle contained) by the whole (straight-line) cutting (the circle), and the (part of it) cut off outside (the circle), between the point and the convex circumference, will be equal to the square on the tangent (line).
(Which is) the very thing it was required to show.
Proposition 37
If some point is taken outside a circle, and two straight-lines radiate from the point towards the circle, and one of them cuts the circle, and the (other) meets (it), and the (rectangle contained) by the whole (straight-line) cutting (the circle), and the (part of it) cut off outside (the circle), between the point and the convex circumference, is equal to the (square) on the (straight-line) meeting (the circle), then the (straight-line) meeting (the circle) will touch the circle.
For let some point D have been taken outside circle ABC, and let two straight-lines, DCA and DB, radiate from D towards circle ABC, and let DCA cut the circle, and let DB meet (the circle).
And let the (rectangle contained) by AD and DC be equal to the (square) on DB.
I say that DB touches circle ABC.
For let DE have been drawn touching ABC [Prop. 3.17], and let the center of the circle ABC have been found, and let it be (at) F.
And let FE, FB, and FD have been joined.
(Angle) FED is thus a right-angle [Prop. 3.18].
And since DE touches circle ABC, and DCA cuts (it), the (rectangle contained) by AD and DC is thus equal to the (square) on DE [Prop. 3.36].
And the (rectangle contained) by AD and DC was also equal to the (square) on DB.
Thus, the (square) on DE is equal to the (square) on DB.
Thus, DE (is) equal to DB.
And FE is also equal to FB.
So the two (straight-lines) DE, EF are equal to the two (straight-lines) DB, BF (respectively).
And their base, FD, is common.
Thus, angle DEF is equal to angle DBF [Prop. 1.8].
And DEF (is) a right-angle.
Thus, DBF (is) also a right-angle.
And FB produced is a diameter, And a (straight-line) drawn at right-angles to a diameter of a circle, at its extremity, touches the circle [Prop. 3.16 corr.].
Thus, DB touches circle ABC.
Similarly, (the same thing) can be shown, even if the center happens to be on AC.
Thus, if some point is taken outside a circle, and two straight-lines radiate from the point towards the circle, and one of them cuts the circle, and the (other) meets (it), and the (rectangle contained) by the whole (straight-line) cutting (the circle), and the (part of it) cut off outside (the circle), between the point and the convex circumference, is equal to the (square) on the (straight-line) meeting (the circle), then the (straight-line) meeting (the circle) will touch the circle.
(Which is) the very thing it was required to show.