1. Similar rectilinear figures are those (which) have (their) angles separately equal and the (corresponding) sides about the equal angles proportional.
2. A straight-line is said to have been cut in extreme and mean ratio when as the whole is to the greater segment so the greater (segment is) to the lesser.
3. The height of any figure is the (straight-line) drawn from the vertex perpendicular to the base.
Proposition 1
Triangles and parallelograms which are of the same height are to one another as their bases.
Let ABC and ACD be triangles, and EC and CF parallelograms, of the same height AC.
I say that as base BC is to base CD, so triangle ABC (is) to triangle ACD, and parallelogram EC to parallelogram CF.
For let the (straight-line) BD have been produced in each direction to points H and L, and let [any number] (of straight-lines) BG and GH be made equal to base BC, and any number (of straight-lines) DK and KL equal to base CD. And let AG, AH, AK, and AL have been joined.
And since CB, BG, and GH are equal to one another, triangles AHG, AGB, and ABC are also equal to one another [Prop. 1.38].
Thus, as many times as base HC is (divisible by) base BC, so many times is triangle AHC also (divisible) by triangle ABC.
So, for the same (reasons), as many times as base LC is (divisible) by base CD, so many times is triangle ALC also (divisible) by triangle ACD.
And if base HC is equal to base CL then triangle AHC is also equal to triangle ACL [Prop. 1.38].
And if base HC exceeds base CL then triangle AHC also exceeds triangle ACL.
And if ( HC is) less (than CL then AHC is also) less (than ACL).
So, their being four magnitudes, two bases, BC and CD, and two triangles, ABC and ACD, equal multiples have been taken of base BC and triangle ABC ---(namely), base HC and triangle AHC ---and other random equal multiples of base CD and triangle ADC ---(namely), base LC and triangle ALC.
And it has been shown that if base HC exceeds base CL then triangle AHC also exceeds triangle ALC, and if ( HC is) equal (to CL then AHC is also) equal (to ALC), and if ( HC is) less (than CL then AHC is also) less (than ALC).
Thus, as base BC is to base CD, so triangle ABC (is) to triangle ACD [Def. 5.5].
And since parallelogram EC is double triangle ABC, and parallelogram FC is double triangle ACD [Prop. 1.34], and parts have the same ratio as similar multiples [Prop. 5.15], thus as triangle ABC is to triangle ACD, so parallelogram EC (is) to parallelogram FC.
In fact, since it was shown that as base BC (is) to CD, so triangle ABC (is) to triangle ACD, and as triangle ABC (is) to triangle ACD, so parallelogram EC (is) to parallelogram CF, thus, also, as base BC (is) to base CD, so parallelogram EC (is) to parallelogram FC [Prop. 5.11].
Thus, triangles and parallelograms which are of the same height are to one another as their bases.
(Which is) the very thing it was required to show.
Proposition 2
If some straight-line is drawn parallel to one of the sides of a triangle then it will cut the (other) sides of the triangle proportionally.
And if (two of) the sides of a triangle are cut proportionally then the straight-line joining the cutting (points) will be parallel to the remaining side of the triangle.
For let DE have been drawn parallel to one of the sides BC of triangle ABC.
I say that as BD is to DA, so CE (is) to EA.
For let BE and CD have been joined.
Thus, triangle BDE is equal to triangle CDE.
For they are on the same base DE and between the same parallels DE and BC [Prop. 1.38].
And ADE is some other triangle.
And equal (magnitudes) have the same ratio to the same (magnitude) [Prop. 5.7].
Thus, as triangle BDE is to [triangle] ADE, so triangle CDE (is) to triangle ADE.
But, as triangle BDE (is) to triangle ADE, so (is) BD to DA.
For, having the same height---(namely), the (straight-line) drawn from E perpendicular to AB --- they are to one another as their bases [Prop. 6.1].
So, for the same (reasons), as triangle CDE (is) to ADE, so CE (is) to EA.
And, thus, as BD (is) to DA, so CE (is) to EA [Prop. 5.11].
And so, let the sides AB and AC of triangle ABC have been cut proportionally (such that) as BD (is) to DA, so CE (is) to EA.
And let DE have been joined.
I say that DE is parallel to BC.
For, by the same construction, since as BD is to DA, so CE (is) to EA, but as BD (is) to DA, so triangle BDE (is) to triangle ADE, and as CE (is) to EA, so triangle CDE (is) to triangle ADE [Prop. 6.1], thus, also, as triangle BDE (is) to triangle ADE, so triangle CDE (is) to triangle ADE [Prop. 5.11].
Thus, triangles BDE and CDE each have the same ratio to ADE.
Thus, triangle BDE is equal to triangle CDE [Prop. 5.9].
And they are on the same base DE.
And equal triangles, which are also on the same base, are also between the same parallels [Prop. 1.39].
Thus, DE is parallel to BC.
Thus, if some straight-line is drawn parallel to one of the sides of a triangle, then it will cut the (other) sides of the triangle proportionally.
And if (two of) the sides of a triangle are cut proportionally, then the straight-line joining the cutting (points) will be parallel to the remaining side of the triangle.
(Which is) the very thing it was required to show.
Proposition 3
If an angle of a triangle is cut in half, and the straight-line cutting the angle also cuts the base, then the segments of the base will have the same ratio as the remaining sides of the triangle.
And if the segments of the base have the same ratio as the remaining sides of the triangle, then the straight-line joining the vertex to the cutting (point) will cut the angle of the triangle in half.
Let ABC be a triangle.
And let the angle BAC have been cut in half by the straight-line AD.
I say that as BD is to CD, so BA (is) to AC.
For let CE have been drawn through (point) C parallel to DA.
And, BA being drawn through, let it meet ( CE) at (point) E.
And since the straight-line AC falls across the parallel (straight-lines) AD and EC, angle ACE is thus equal to CAD [Prop. 1.29].
But, (angle) CAD is assumed (to be) equal to BAD.
Thus, (angle) BAD is also equal to ACE.
Again, since the straight-line BAE falls across the parallel (straight-lines) AD and EC, the external angle BAD is equal to the internal (angle) AEC [Prop. 1.29].
And (angle) ACE was also shown (to be) equal to BAD.
Thus, angle ACE is also equal to AEC.
And, hence, side AE is equal to side AC [Prop. 1.6].
And since AD has been drawn parallel to one of the sides EC of triangle BCE, thus, proportionally, as BD is to DC, so BA (is) to AE [Prop. 6.2].
And AE (is) equal to AC.
Thus, as BD (is) to DC, so BA (is) to AC.
And so, let BD be to DC, as BA (is) to AC.
And let AD have been joined.
I say that angle BAC has been cut in half by the straight-line AD.
For, by the same construction, since as BD is to DC, so BA (is) to AC, then also as BD (is) to DC, so BA is to AE.
For AD has been drawn parallel to one (of the sides) EC of triangle BCE [Prop. 6.2].
Thus, also, as BA (is) to AC, so BA (is) to AE [Prop. 5.11].
Thus, AC (is) equal to AE [Prop. 5.9].
And, hence, angle AEC is equal to ACE [Prop. 1.5].
But, AEC [is] equal to the external (angle) BAD, and ACE is equal to the alternate (angle) CAD [Prop. 1.29].
Thus, (angle) BAD is also equal to CAD.
Thus, angle BAC has been cut in half by the straight-line AD.
Thus, if an angle of a triangle is cut in half, and the straight-line cutting the angle also cuts the base, then the segments of the base will have the same ratio as the remaining sides of the triangle.
And if the segments of the base have the same ratio as the remaining sides of the triangle, then the straight-line joining the vertex to the cutting (point) will cut the angle of the triangle in half.
(Which is) the very thing it was required to show.
Proposition 4
In equiangular triangles the sides about the equal angles are proportional, and those (sides) subtending equal angles correspond.
Let ABC and DCE be equiangular triangles, having angle ABC equal to DCE, and (angle) BAC to CDE, and, further, (angle) ACB to CED.
I say that in triangles ABC and DCE the sides about the equal angles are proportional, and those (sides) subtending equal angles correspond.
Let BC be placed straight-on to CE.
And since angles ABC and ACB are less than two right-angles [Prop. 1.17], and ACB (is) equal to DEC, thus ABC and DEC are less than two right-angles.
Thus, BA and ED, being produced, will meet [C.N. 5].
Let them have been produced, and let them meet at (point) F.
And since angle DCE is equal to ABC, BF is parallel to CD [Prop. 1.28].
Again, since (angle) ACB is equal to DEC, AC is parallel to FE [Prop. 1.28].
Thus, FACD is a parallelogram.
Thus, FA is equal to DC, and AC to FD [Prop. 1.34].
And since AC has been drawn parallel to one (of the sides) FE of triangle FBE, thus as BA is to AF, so BC (is) to CE [Prop. 6.2].
And AF (is) equal to CD.
Thus, as BA (is) to CD, so BC (is) to CE, and, alternately, as AB (is) to BC, so DC (is) to CE [Prop. 5.16].
Again, since CD is parallel to BF, thus as BC (is) to CE, so FD (is) to DE [Prop. 6.2].
And FD (is) equal to AC.
Thus, as BC is to CE, so AC (is) to DE, and, alternately, as BC (is) to CA, so CE (is) to ED [Prop. 6.2].
Therefore, since it was shown that as AB (is) to BC, so DC (is) to CE, and as BC (is) to CA, so CE (is) to ED, thus, via equality, as BA (is) to AC, so CD (is) to DE [Prop. 5.22].
Thus, in equiangular triangles the sides about the equal angles are proportional, and those (sides) subtending equal angles correspond.
(Which is) the very thing it was required to show.
Proposition 5
If two triangles have proportional sides then the triangles will be equiangular, and will have the angles which corresponding sides subtend equal.
Let ABC and DEF be two triangles having proportional sides, (so that) as AB (is) to BC, so DE (is) to EF, and as BC (is) to CA, so EF (is) to FD, and, further, as BA (is) to AC, so ED (is) to DF.
I say that triangle ABC is equiangular to triangle DEF, and (that the triangles) will have the angles which corresponding sides subtend equal.
(That is), (angle) ABC (equal) to DEF, BCA to EFD, and, further, BAC to EDF.
For let (angle) FEG, equal to angle ABC, and (angle) EFG, equal to ACB, have been constructed on the straight-line EF at the points E and F on it (respectively) [Prop. 1.23].
Thus, the remaining (angle) at A is equal to the remaining (angle) at G [Prop. 1.32].
Thus, triangle ABC is equiangular to [triangle] EGF.
Thus, for triangles ABC and EGF, the sides about the equal angles are proportional, and (those) sides subtending equal angles correspond [Prop. 6.4].
Thus, as AB is to BC, [so] GE (is) to EF.
But, as AB (is) to BC, so, it was assumed, (is) DE to EF.
Thus, as DE (is) to EF, so GE (is) to EF [Prop. 5.11].
Thus, DE and GE each have the same ratio to EF.
Thus, DE is equal to GE [Prop. 5.9].
So, for the same (reasons), DF is also equal to GF.
Therefore, since DE is equal to EG, and EF (is) common, the two (sides) DE, EF are equal to the two (sides) GE, EF (respectively).
And base DF [is] equal to base FG.
Thus, angle DEF is equal to angle GEF [Prop. 1.8], and triangle DEF (is) equal to triangle GEF, and the remaining angles (are) equal to the remaining angles which the equal sides subtend [Prop. 1.4].
Thus, angle DFE is also equal to GFE, and (angle) EDF to EGF.
And since (angle) FED is equal to GEF, and (angle) GEF to ABC, angle ABC is thus also equal to DEF.
So, for the same (reasons), (angle) ACB is also equal to DFE, and, further, the (angle) at A to the (angle) at D.
Thus, triangle ABC is equiangular to triangle DEF.
Thus, if two triangles have proportional sides then the triangles will be equiangular, and will have the angles which corresponding sides subtend equal.
(Which is) the very thing it was required to show.
Proposition 6
If two triangles have one angle equal to one angle, and the sides about the equal angles proportional, then the triangles will be equiangular, and will have the angles which corresponding sides subtend equal.
Let ABC and DEF be two triangles having one angle, BAC, equal to one angle, EDF (respectively), and the sides about the equal angles proportional, (so that) as BA (is) to AC, so ED (is) to DF.
I say that triangle ABC is equiangular to triangle DEF, and will have angle ABC equal to DEF, and (angle) ACB to DFE.
For let (angle) FDG, equal to each of BAC and EDF, and (angle) DFG, equal to ACB, have been constructed on the straight-line DF at the points D and F on it (respectively) [Prop. 1.23].
Thus, the remaining angle at B is equal to the remaining angle at G [Prop. 1.32].
Thus, triangle ABC is equiangular to triangle DGF.
Thus, proportionally, as BA (is) to AC, so GD (is) to DF [Prop. 6.4].
And it was also assumed that as BA is) to AC, so ED (is) to DF.
And, thus, as ED (is) to DF, so GD (is) to DF [Prop. 5.11].
Thus, ED (is) equal to DG [Prop. 5.9].
And DF (is) common.
So, the two (sides) ED, DF are equal to the two (sides) GD, DF (respectively).
And angle EDF [is] equal to angle GDF.
Thus, base EF is equal to base GF, and triangle DEF is equal to triangle GDF, and the remaining angles will be equal to the remaining angles which the equal sides subtend [Prop. 1.4].
Thus, (angle) DFG is equal to DFE, and (angle) DGF to DEF.
But, (angle) DFG is equal to ACB.
Thus, (angle) ACB is also equal to DFE.
And (angle) BAC was also assumed (to be) equal to EDF.
Thus, the remaining (angle) at B is equal to the remaining (angle) at E [Prop. 1.32].
Thus, triangle ABC is equiangular to triangle DEF.
Thus, if two triangles have one angle equal to one angle, and the sides about the equal angles proportional, then the triangles will be equiangular, and will have the angles which corresponding sides subtend equal.
(Which is) the very thing it was required to show.
Proposition 7
If two triangles have one angle equal to one angle, and the sides about other angles proportional, and the remaining angles either both less than, or both not less than, right-angles, then the triangles will be equiangular, and will have the angles about which the sides are proportional equal.
Let ABC and DEF be two triangles having one angle, BAC, equal to one angle, EDF (respectively), and the sides about (some) other angles, ABC and DEF (respectively), proportional, (so that) as AB (is) to BC, so DE (is) to EF, and the remaining (angles) at C and F, first of all, both less than right-angles.
I say that triangle ABC is equiangular to triangle DEF, and (that) angle ABC will be equal to DEF, and (that) the remaining (angle) at C (will be) manifestly equal to the remaining (angle) at F.
For if angle ABC is not equal to (angle) DEF then one of them is greater.
Let ABC be greater.
And let (angle) ABG, equal to (angle) DEF, have been constructed on the straight-line AB at the point B on it [Prop. 1.23].
And since angle BAG is equal to (angle) EDF, and (angle) ABG to DEF, the remaining (angle) AGB is thus equal to the remaining (angle) DFE [Prop. 1.32].
Thus, triangle ABG is equiangular to triangle DEF.
Thus, as AB is to BG, so DE (is) to EF [Prop. 6.4].
And as DE (is) to EF, [so] it was assumed (is) AB to BC.
Thus, AB has the same ratio to each of BC and BG [Prop. 5.11].
Thus, BC (is) equal to BG [Prop. 5.9].
And, hence, the angle at C is equal to angle BGC [Prop. 1.5].
And the angle at C was assumed (to be) less than a right-angle.
Thus, (angle) BGC is also less than a right-angle.
Hence, the adjacent angle to it, AGB, is greater than a right-angle [Prop. 1.13].
And ( AGB) was shown to be equal to the (angle) at F.
Thus, the (angle) at F is also greater than a right-angle.
But it was assumed (to be) less than a right-angle.
The very thing is absurd.
Thus, angle ABC is not unequal to (angle) DEF.
Thus, (it is) equal.
And the (angle) at A is also equal to the (angle) at D.
And thus the remaining (angle) at C is equal to the remaining (angle) at F [Prop. 1.32].
Thus, triangle ABC is equiangular to triangle DEF.
But, again, let each of the (angles) at C and F be assumed (to be) not less than a right-angle.
I say, again, that triangle ABC is equiangular to triangle DEF in this case also.
For, with the same construction, we can similarly show that BC is equal to BG.
Hence, also, the angle at C is equal to (angle) BGC.
And the (angle) at C (is) not less than a right-angle.
Thus, BGC (is) not less than a right-angle either.
So, in triangle BGC the (sum of) two angles is not less than two right-angles.
The very thing is impossible [Prop. 1.17].
Thus, again, angle ABC is not unequal to DEF.
Thus, (it is) equal.
And the (angle) at A is also equal to the (angle) at D.
Thus, the remaining (angle) at C is equal to the remaining (angle) at F [Prop. 1.32].
Thus, triangle ABC is equiangular to triangle DEF.
Thus, if two triangles have one angle equal to one angle, and the sides about other angles proportional, and the remaining angles both less than, or both not less than, right-angles, then the triangles will be equiangular, and will have the angles about which the sides (are) proportional equal.
(Which is) the very thing it was required to show.
Proposition 8
If, in a right-angled triangle, a (straight-line) is drawn from the right-angle perpendicular to the base then the triangles around the perpendicular are similar to the whole (triangle), and to one another.
Let ABC be a right-angled triangle having the angle BAC a right-angle, and let AD have been drawn from A, perpendicular to BC [Prop. 1.12].
I say that triangles ABD and ADC are each similar to the whole (triangle) ABC and, further, to one another.
For since (angle) BAC is equal to ADB ---for each (are) right-angles---and the (angle) at B (is) common to the two triangles ABC and ABD, the remaining (angle) ACB is thus equal to the remaining (angle) BAD [Prop. 1.32].
Thus, triangle ABC is equiangular to triangle ABD.
Thus, as BC, subtending the right-angle in triangle ABC, is to BA, subtending the right-angle in triangle ABD, so the same AB, subtending the angle at C in triangle ABC, (is) to BD, subtending the equal (angle) BAD in triangle ABD, and, further, (so is) AC to AD, (both) subtending the angle at B common to the two triangles [Prop. 6.4].
Thus, triangle ABC is equiangular to triangle ABD, and has the sides about the equal angles proportional.
Thus, triangle ABC [is] similar to triangle ABD [Def. 6.1].
So, similarly, we can show that triangle ABC is also similar to triangle ADC.
Thus, [triangles] ABD and ADC are each similar to the whole (triangle) ABC.
So I say that triangles ABD and ADC are also similar to one another.
For since the right-angle BDA is equal to the right-angle ADC, and, indeed, (angle) BAD was also shown (to be) equal to the (angle) at C, thus the remaining (angle) at B is also equal to the remaining (angle) DAC [Prop. 1.32].
Thus, triangle ABD is equiangular to triangle ADC.
Thus, as BD, subtending (angle) BAD in triangle ABD, is to DA, subtending the (angle) at C in triangle ADC, (which is) equal to (angle) BAD, so (is) the same AD, subtending the angle at B in triangle ABD, to DC, subtending (angle) DAC in triangle ADC, (which is) equal to the (angle) at B, and, further, (so is) BA to AC, (each) subtending right-angles [Prop. 6.4].
Thus, triangle ABD is similar to triangle ADC [Def. 6.1].
Thus, if, in a right-angled triangle, a (straight-line) is drawn from the right-angle perpendicular to the base then the triangles around the perpendicular are similar to the whole (triangle), and to one another.
[(Which is) the very thing it was required to show.]
Corollary
So (it is) clear, from this, that if, in a right-angled triangle, a (straight-line) is drawn from the right-angle perpendicular to the base then the (straight-line so) drawn is in mean proportion to the pieces of the base.
(which is) the very thing it was required to show.
Proposition 9
To cut off a prescribed part from a given straight-line.
Let AB be the given straight-line.
So it is required to cut off a prescribed part from AB.
So let a third (part) have been prescribed.
[And] let some straight-line AC have been drawn from (point) A, encompassing a random angle with AB.
And let a random point D have been taken on AC.
And let DE and EC be made equal to AD [Prop. 1.3].
And let BC have been joined.
And let DF have been drawn through D parallel to it [Prop. 1.31].
Therefore, since FD has been drawn parallel to one of the sides, BC, of triangle ABC, then, proportionally, as CD is to DA, so BF (is) to FA [Prop. 6.2].
And CD (is) double DA.
Thus, BF (is) also double FA.
Thus, BA (is) triple AF.
Thus, the prescribed third part, AF, has been cut off from the given straight-line, AB.
(Which is) the very thing it was required to do.
Proposition 10
To cut a given uncut straight-line similarly to a given cut (straight-line).
Let AB be the given uncut straight-line, and AC a (straight-line) cut at points D and E, and let ( AC) be laid down so as to encompass a random angle (with AB).
And let CB have been joined.
And let DF and EG have been drawn through (points) D and E (respectively), parallel to BC, and let DHK have been drawn through (point) D, parallel to AB [Prop. 1.31].
Thus, FH and HB are each parallelograms.
Thus, DH (is) equal to FG, and HK to GB [Prop. 1.34].
And since the straight-line HE has been drawn parallel to one of the sides, KC, of triangle DKC, thus, proportionally, as CE is to ED, so KH (is) to HD [Prop. 6.2].
And KH (is) equal to BG, and HD to GF.
Thus, as CE is to ED, so BG (is) to GF.
Again, since FD has been drawn parallel to one of the sides, GE, of triangle AGE, thus, proportionally, as ED is to DA, so GF (is) to FA [Prop. 6.2].
And it was also shown that as CE (is) to ED, so BG (is) to GF.
Thus, as CE is to ED, so BG (is) to GF, and as ED (is) to DA, so GF (is) to FA.
Thus, the given uncut straight-line, AB, has been cut similarly to the given cut straight-line, AC.
(Which is) the very thing it was required to do.
Proposition 11
To find a third (straight-line) proportional to two given straight-lines.
Let BA and AC be the [two] given [straight-lines], and let them be laid down encompassing a random angle.
So it is required to find a third (straight-line) proportional to BA and AC.
For let ( BA and AC) have been produced to points D and E (respectively), and let BD be made equal to AC [Prop. 1.3].
And let BC have been joined.
And let DE have been drawn through (point) D parallel to it [Prop. 1.31].
Therefore, since BC has been drawn parallel to one of the sides DE of triangle ADE, proportionally, as AB is to BD, so AC (is) to CE [Prop. 6.2].
And BD (is) equal to AC.
Thus, as AB is to AC, so AC (is) to CE.
Thus, a third (straight-line), CE, has been found (which is) proportional to the two given straight-lines, AB and AC.
(Which is) the very thing it was required to do.
Proposition 12
To find a fourth (straight-line) proportional to three given straight-lines.
Let A, B, and C be the three given straight-lines.
So it is required to find a fourth (straight-line) proportional to A, B, and C.
Let the two straight-lines DE and DF be set out encompassing the [random] angle EDF.
And let DG be made equal to A, and GE to B, and, further, DH to C [Prop. 1.3].
And GH being joined, let EF have been drawn through (point) E parallel to it [Prop. 1.31].
Therefore, since GH has been drawn parallel to one of the sides EF of triangle DEF, thus as DG is to GE, so DH (is) to HF [Prop. 6.2].
And DG (is) equal to A, and GE to B, and DH to C.
Thus, as A is to B, so C (is) to HF.
Thus, a fourth (straight-line), HF, has been found (which is) proportional to the three given straight-lines, A, B, and C.
(Which is) the very thing it was required to do.
Proposition 13
To find the (straight-line) in mean proportion to two given straight-lines.
Let AB and BC be the two given straight-lines.
So it is required to find the (straight-line) in mean proportion to AB and BC.
Let ( AB and BC) be laid down straight-on (with respect to one another), and let the semi-circle ADC have been drawn on AC [Prop. 1.10].
And let BD have been drawn from (point) B, at right-angles to AC [Prop. 1.11].
And let AD and DC have been joined.
And since ADC is an angle in a semi-circle, it is a right-angle [Prop. 3.31].
And since, in the right-angled triangle ADC, the (straight-line) DB has been drawn from the right-angle perpendicular to the base, DB is thus the mean proportional to the pieces of the base, AB and BC [Prop. 6.8 corr.].
Thus, DB has been found (which is) in mean proportion to the two given straight-lines, AB and BC.
(Which is) the very thing it was required to do.
Proposition 14
In equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional.
And those equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal.
Let AB and BC be equal and equiangular parallelograms having the angles at B equal.
And let DB and BE be laid down straight-on (with respect to one another).
Thus, FB and BG are also straight-on (with respect to one another) [Prop. 1.14].
I say that the sides of AB and BC about the equal angles are reciprocally proportional, that is to say, that as DB is to BE, so GB (is) to BF.
For let the parallelogram FE have been completed.
Therefore, since parallelogram AB is equal to parallelogram BC, and FE (is) some other (parallelogram), thus as (parallelogram) AB is to FE, so (parallelogram) BC (is) to FE [Prop. 5.7].
But, as (parallelogram) AB (is) to FE, so DB (is) to BE, and as (parallelogram) BC (is) to FE, so GB (is) to BF [Prop. 6.1].
Thus, also, as DB (is) to BE, so GB (is) to BF.
Thus, in parallelograms AB and BC the sides about the equal angles are reciprocally proportional.
And so, let DB be to BE, as GB (is) to BF. I say that parallelogram AB is equal to parallelogram BC.
For since as DB is to BE, so GB (is) to BF, but as DB (is) to BE, so parallelogram AB (is) to parallelogram FE, and as GB (is) to BF, so parallelogram BC (is) to parallelogram FE [Prop. 6.1], thus, also, as (parallelogram) AB (is) to FE, so (parallelogram) BC (is) to FE [Prop. 5.11].
Thus, parallelogram AB is equal to parallelogram BC [Prop. 5.9].
Thus, in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional.
And those equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal.
(Which is) the very thing it was required to show.
Proposition 15
In equal triangles also having one angle equal to one (angle) the sides about the equal angles are reciprocally proportional.
And those triangles having one angle equal to one angle for which the sides about the equal angles (are) reciprocally proportional are equal.
Let ABC and ADE be equal triangles having one angle equal to one (angle), (namely) BAC (equal) to DAE.
I say that, in triangles ABC and ADE, the sides about the equal angles are reciprocally proportional, that is to say, that as CA is to AD, so EA (is) to AB.
For let CA be laid down so as to be straight-on (with respect) to AD.
Thus, EA is also straight-on (with respect) to AB [Prop. 1.14].
And let BD have been joined.
Therefore, since triangle ABC is equal to triangle ADE, and BAD (is) some other (triangle), thus as triangle CAB is to triangle BAD, so triangle EAD (is) to triangle BAD [Prop. 5.7].
But, as (triangle) CAB (is) to BAD, so CA (is) to AD, and as (triangle) EAD (is) to BAD, so EA (is) to AB [Prop. 6.1].
And thus, as CA (is) to AD, so EA (is) to AB.
Thus, in triangles ABC and ADE the sides about the equal angles (are) reciprocally proportional.
And so, let the sides of triangles ABC and ADE be reciprocally proportional, and (thus) let CA be to AD, as EA (is) to AB.
I say that triangle ABC is equal to triangle ADE.
For, BD again being joined, since as CA is to AD, so EA (is) to AB, but as CA (is) to AD, so triangle ABC (is) to triangle BAD, and as EA (is) to AB, so triangle EAD (is) to triangle BAD [Prop. 6.1], thus as triangle ABC (is) to triangle BAD, so triangle EAD (is) to triangle BAD.
Thus, (triangles) ABC and EAD each have the same ratio to BAD.
Thus, [triangle] ABC is equal to triangle EAD [Prop. 5.9].
Thus, in equal triangles also having one angle equal to one (angle) the sides about the equal angles (are) reciprocally proportional.
And those triangles having one angle equal to one angle for which the sides about the equal angles (are) reciprocally proportional are equal.
(Which is) the very thing it was required to show.
Proposition 16
If four straight-lines are proportional then the rectangle contained by the (two) outermost is equal to the rectangle contained by the middle (two).
And if the rectangle contained by the (two) outermost is equal to the rectangle contained by the middle (two) then the four straight-lines will be proportional.
Let AB, CD, E, and F be four proportional straight-lines, (such that) as AB (is) to CD, so E (is) to F.
I say that the rectangle contained by AB and F is equal to the rectangle contained by CD and E.
[For] let AG and CH have been drawn from points A and C at right-angles to the straight-lines AB and CD (respectively) [Prop. 1.11].
And let AG be made equal to F, and CH to E [Prop. 1.3].
And let the parallelograms BG and DH have been completed.
And since as AB is to CD, so E (is) to F, and E (is) equal CH, and F to AG, thus as AB is to CD, so CH (is) to AG. Thus, in the parallelograms BG and DH the sides about the equal angles are reciprocally proportional.
And those equiangular parallelograms in which the sides about the equal angles are reciprocally proportional are equal [Prop. 6.14].
Thus, parallelogram BG is equal to parallelogram DH.
And BG is the (rectangle contained) by AB and F.
For AG (is) equal to F.
And DH (is) the (rectangle contained) by CD and E.
For E (is) equal to CH.
Thus, the rectangle contained by AB and F is equal to the rectangle contained by CD and E.
And so, let the rectangle contained by AB and F be equal to the rectangle contained by CD and E.
I say that the four straight-lines will be proportional, (so that) as AB (is) to CD, so E (is) to F.
For, with the same construction, since the (rectangle contained) by AB and F is equal to the (rectangle contained) by CD and E.
And BG is the (rectangle contained) by AB and F.
For AG is equal to F.
And DH (is) the (rectangle contained) by CD and E.
For CH (is) equal to E.
BG is thus equal to DH.
And they are equiangular.
And in equal and equiangular parallelograms the sides about the equal angles are reciprocally proportional [Prop. 6.14].
Thus, as AB is to CD, so CH (is) to AG.
And CH (is) equal to E, and AG to F.
Thus, as AB is to CD, so E (is) to F.
Thus, if four straight-lines are proportional then the rectangle contained by the (two) outermost is equal to the rectangle contained by the middle (two).
And if the rectangle contained by the (two) outermost is equal to the rectangle contained by the middle (two) then the four straight-lines will be proportional.
(Which is) the very thing it was required to show.
Proposition 17
If three straight-lines are proportional then the rectangle contained by the (two) outermost is equal to the square on the middle (one).
And if the rectangle contained by the (two) outermost is equal to the square on the middle (one) then the three straight-lines will be proportional.
Let A, B and C be three proportional straight-lines, (such that) as A (is) to B, so B (is) to C.
I say that the rectangle contained by A and C is equal to the square on B.
Let D be made equal to B [Prop. 1.3].
And since as A is to B, so B (is) to C, and B (is) equal to D, thus as A is to B, (so) D (is) to C.
And if four straight-lines are proportional then the [rectangle] contained by the (two) outermost is equal to the rectangle contained by the middle (two) [Prop. 6.16].
Thus, the (rectangle contained) by A and C is equal to the (rectangle contained) by B and D.
But, the (rectangle contained) by B and D is the (square) on B.
For B (is) equal to D.
Thus, the rectangle contained by A and C is equal to the square on B.
And so, let the (rectangle contained) by A and C be equal to the (square) on B.
I say that as A is to B, so B (is) to C.
For, with the same construction, since the (rectangle contained) by A and C is equal to the (square) on B.
But, the (square) on B is the (rectangle contained) by B and D.
For B (is) equal to D.
The (rectangle contained) by A and C is thus equal to the (rectangle contained) by B and D.
And if the (rectangle contained) by the (two) outermost is equal to the (rectangle contained) by the middle (two) then the four straight-lines are proportional [Prop. 6.16].
Thus, as A is to B, so D (is) to C.
And B (is) equal to D.
Thus, as A (is) to B, so B (is) to C.
Thus, if three straight-lines are proportional then the rectangle contained by the (two) outermost is equal to the square on the middle (one).
And if the rectangle contained by the (two) outermost is equal to the square on the middle (one) then the three straight-lines will be proportional.
(Which is) the very thing it was required to show.
Proposition 18
To describe a rectilinear figure similar, and similarly laid down, to a given rectilinear figure on a given straight-line.
Let AB be the given straight-line, and CE the given rectilinear figure.
So it is required to describe a rectilinear figure similar, and similarly laid down, to the rectilinear figure CE on the straight-line AB.
Let DF have been joined, and let GAB, equal to the angle at C, and ABG, equal to (angle) CDF, have been constructed on the straight-line AB at the points A and B on it (respectively) [Prop. 1.23].
Thus, the remaining (angle) CFD is equal to AGB [Prop. 1.32].
Thus, triangle FCD is equiangular to triangle GAB.
Thus, proportionally, as FD is to GB, so FC (is) to GA, and CD to AB [Prop. 6.4].
Again, let BGH, equal to angle DFE, and GBH equal to (angle) FDE, have been constructed on the straight-line BG at the points G and B on it (respectively) [Prop. 1.23].
Thus, the remaining (angle) at E is equal to the remaining (angle) at H [Prop. 1.32].
Thus, triangle FDE is equiangular to triangle GHB.
Thus, proportionally, as FD is to GB, so FE (is) to GH, and ED to HB [Prop. 6.4].
And it was also shown (that) as FD (is) to GB, so FC (is) to GA, and CD to AB.
Thus, also, as FC (is) to AG, so CD (is) to AB, and FE to GH, and, further, ED to HB.
And since angle CFD is equal to AGB, and DFE to BGH, thus the whole (angle) CFE is equal to the whole (angle) AGH.
So, for the same (reasons), (angle) CDE is also equal to ABH.
And the (angle) at C is also equal to the (angle) at A, and the (angle) at E to the (angle) at H.
Thus, (figure) AH is equiangular to CE.
And (the two figures) have the sides about their equal angles proportional.
Thus, the rectilinear figure AH is similar to the rectilinear figure CE [Def. 6.1].
Thus, the rectilinear figure AH, similar, and similarly laid down, to the given rectilinear figure CE has been constructed on the given straight-line AB.
(Which is) the very thing it was required to do.
Proposition 19
Similar triangles are to one another in the squared ratio of (their) corresponding sides.
Let ABC and DEF be similar triangles having the angle at B equal to the (angle) at E, and AB to BC, as DE (is) to EF, such that BC corresponds to EF.
I say that triangle ABC has a squared ratio to triangle DEF with respect to (that side) BC (has) to EF.
For let a third (straight-line), BG, have been taken (which is) proportional to BC and EF, so that as BC (is) to EF, so EF (is) to BG [Prop. 6.11].
And let AG have been joined.
Therefore, since as AB is to BC, so DE (is) to EF, thus, alternately, as AB is to DE, so BC (is) to EF [Prop. 5.16].
But, as BC (is) to EF, so EF is to BG.
And, thus, as AB (is) to DE, so EF (is) to BG.
Thus, for triangles ABG and DEF, the sides about the equal angles are reciprocally proportional.
And those triangles having one (angle) equal to one (angle) for which the sides about the equal angles are reciprocally proportional are equal [Prop. 6.15].
Thus, triangle ABG is equal to triangle DEF.
And since as BC (is) to EF, so EF (is) to BG, and if three straight-lines are proportional then the first has a squared ratio to the third with respect to the second [Def. 5.9], BC thus has a squared ratio to BG with respect to (that) CB (has) to EF.
And as CB (is) to BG, so triangle ABC (is) to triangle ABG [Prop. 6.1].
Thus, triangle ABC also has a squared ratio to (triangle) ABG with respect to (that side) BC (has) to EF.
And triangle ABG (is) equal to triangle DEF.
Thus, triangle ABC also has a squared ratio to triangle DEF with respect to (that side) BC (has) to EF.
Thus, similar triangles are to one another in the squared ratio of (their) corresponding sides.
[(Which is) the very thing it was required to show].
Corollary
So it is clear, from this, that if three straight-lines are proportional, then as the first is to the third, so the figure (described) on the first (is) to the similar, and similarly described, (figure) on the second.
(Which is) the very thing it was required to show.
Proposition 20
Similar polygons can be divided into equal numbers of similar triangles corresponding (in proportion) to the wholes, and one polygon has to the (other) polygon a squared ratio with respect to (that) a corresponding side (has) to a corresponding side.
Let ABCDE and FGHKL be similar polygons, and let AB correspond to FG.
I say that polygons ABCDE and FGHKL can be divided into equal numbers of similar triangles corresponding (in proportion) to the wholes, and (that) polygon ABCDE has a squared ratio to polygon FGHKL with respect to that AB (has) to FG.
Let BE, EC, GL, and LH have been joined.
And since polygon ABCDE is similar to polygon FGHKL, angle BAE is equal to angle GFL, and as BA is to AE, so GF (is) to FL [Def. 6.1].
Therefore, since ABE and FGL are two triangles having one angle equal to one angle and the sides about the equal angles proportional, triangle ABE is thus equiangular to triangle FGL [Prop. 6.6].
Hence, (they are) also similar [Prop. 6.4] [Def. 6.1].
Thus, angle ABE is equal to (angle) FGL.
And the whole (angle) ABC is equal to the whole (angle) FGH, on account of the similarity of the polygons.
Thus, the remaining angle EBC is equal to LGH.
And since, on account of the similarity of triangles ABE and FGL, as EB is to BA, so LG (is) to GF, but also, on account of the similarity of the polygons, as AB is to BC, so FG (is) to GH, thus, via equality, as EB is to BC, so LG (is) to GH [Prop. 5.22], and the sides about the equal angles, EBC and LGH, are proportional.
Thus, triangle EBC is equiangular to triangle LGH [Prop. 6.6].
Hence, triangle EBC is also similar to triangle LGH [Prop. 6.4] [Def. 6.1].
So, for the same (reasons), triangle ECD is also similar to triangle LHK.
Thus, the similar polygons ABCDE and FGHKL have been divided into equal numbers of similar triangles.
I also say that (the triangles) correspond (in proportion) to the wholes.
That is to say, the triangles are proportional: ABE, EBC, and ECD are the leading (magnitudes), and their (associated) following (magnitudes are) FGL, LGH, and LHK (respectively).
(I) also (say) that polygon ABCDE has a squared ratio to polygon FGHKL with respect to (that) a corresponding side (has) to a corresponding side---that is to say, (side) AB to FG.
For let AC and FH have been joined.
And since angle ABC is equal to FGH, and as AB is to BC, so FG (is) to GH, on account of the similarity of the polygons, triangle ABC is equiangular to triangle FGH [Prop. 6.6].
Thus, angle BAC is equal to GFH, and (angle) BCA to GHF.
And since angle BAM is equal to GFN, and (angle) ABM is also equal to FGN (see earlier), the remaining (angle) AMB is thus also equal to the remaining (angle) FNG [Prop. 1.32].
Thus, triangle ABM is equiangular to triangle FGN.
So, similarly, we can show that triangle BMC is also equiangular to triangle GNH.
Thus, proportionally, as AM is to MB, so FN (is) to NG, and as BM (is) to MC, so GN (is) to NH [Prop. 6.4].
Hence, also, via equality, as AM (is) to MC, so FN (is) to NH [Prop. 5.22].
But, as AM (is) to MC, so [triangle] ABM is to MBC, and AME to EMC.
For they are to one another as their bases [Prop. 6.1].
And as one of the leading (magnitudes) is to one of the following (magnitudes), so (the sum of) all the leading (magnitudes) is to (the sum of) all the following (magnitudes) [Prop. 5.12].
Thus, as triangle AMB (is) to BMC, so (triangle) ABE (is) to CBE.
But, as (triangle) AMB (is) to BMC, so AM (is) to MC.
Thus, also, as AM (is) to MC, so triangle ABE (is) to triangle EBC.
And so, for the same (reasons), as FN (is) to NH, so triangle FGL (is) to triangle GLH.
And as AM is to MC, so FN (is) to NH.
Thus, also, as triangle ABE (is) to triangle BEC, so triangle FGL (is) to triangle GLH, and, alternately, as triangle ABE (is) to triangle FGL, so triangle BEC (is) to triangle GLH [Prop. 5.16].
So, similarly, we can also show, by joining BD and GK, that as triangle BEC (is) to triangle LGH, so triangle ECD (is) to triangle LHK.
And since as triangle ABE is to triangle FGL, so (triangle) EBC (is) to LGH, and, further, (triangle) ECD to LHK, and also as one of the leading (magnitudes is) to one of the following, so (the sum of) all the leading (magnitudes is) to (the sum of) all the following [Prop. 5.12], thus as triangle ABE is to triangle FGL, so polygon ABCDE (is) to polygon FGHKL.
But, triangle ABE has a squared ratio to triangle FGL with respect to (that) the corresponding side AB (has) to the corresponding side FG.
For, similar triangles are in the squared ratio of corresponding sides [Prop. 6.14].
Thus, polygon ABCDE also has a squared ratio to polygon FGHKL with respect to (that) the corresponding side AB (has) to the corresponding side FG.
Thus, similar polygons can be divided into equal numbers of similar triangles corresponding (in proportion) to the wholes, and one polygon has to the (other) polygon a squared ratio with respect to (that) a corresponding side (has) to a corresponding side.
[(Which is) the very thing it was required to show].
Corollary
And, in the same manner, it can also be shown for [similar] quadrilaterals that they are in the squared ratio of (their) corresponding sides.
And it was also shown for triangles.
Hence, in general, similar rectilinear figures are also to one another in the squared ratio of (their) corresponding sides.
(Which is) the very thing it was required to show.
Proposition 21
(Rectilinear figures) similar to the same rectilinear figure are also similar to one another.
Let each of the rectilinear figures A and B be similar to (the rectilinear figure) C.
I say that A is also similar to B.
For since A is similar to C, ( A) is equiangular to ( C), and has the sides about the equal angles proportional [Def. 6.1].
Again, since B is similar to C, ( B) is equiangular to ( C), and has the sides about the equal angles proportional [Def. 6.1].
Thus, A and B are each equiangular to C, and have the sides about the equal angles proportional.
Thus, A is similar to B [Def. 6.1].
(Which is) the very thing it was required to show.
Proposition 22
If four straight-lines are proportional then similar, and similarly described, rectilinear figures (drawn) on them will also be proportional.
And if similar, and similarly described, rectilinear figures (drawn) on them are proportional then the straight-lines themselves will also 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, rectilinear figures KAB and LCD have been described on AB and CD (respectively), and the similar, and similarly laid out, rectilinear figures MF and NH on EF and GH (respectively).
I say that as KAB is to LCD, so MF (is) to NH.
For let a third (straight-line) O have been taken (which is) proportional to AB and CD, and a third (straight-line) P proportional to EF and GH [Prop. 6.11].
And since as AB is to CD, so EF (is) to GH, and as CD (is) to O, so GH (is) to P, thus, via equality, as AB is to O, so EF (is) to P [Prop. 5.22].
But, as AB (is) to O, so [also] KAB (is) to LCD, and as EF (is) to P, so MF (is) to NH [Prop. 5.19 corr.].
And, thus, as KAB (is) to LCD, so MF (is) to NH.
And so let KAB be to LCD, as MF (is) to NH.
I say also that as AB is to CD, so EF (is) to GH.
For if as AB is to CD, so EF (is) not to GH, let AB be to CD, as EF (is) to QR [Prop. 6.12].
And let the rectilinear figure SR, similar, and similarly laid down, to either of MF or NH, have been described on QR [Prop. 6.18] [Prop. 6.21].
Therefore, since as AB is to CD, so EF (is) to QR, and the similar, and similarly laid out, (rectilinear figures) KAB and LCD have been described on AB and CD (respectively), and the similar, and similarly laid out, (rectilinear figures) MF and SR on EF and QR (resespectively), thus as KAB is to LCD, so MF (is) to SR (see above).
And it was also assumed that as KAB (is) to LCD, so MF (is) to NH.
Thus, also, as MF (is) to SR, so MF (is) to NH [Prop. 5.11].
Thus, MF has the same ratio to each of NH and SR.
Thus, NH is equal to SR [Prop. 5.9].
And it is also similar, and similarly laid out, to it.
Thus, GH (is) equal to QR.
And since AB is to CD, as EF (is) to QR, and QR (is) equal to GH, thus as AB is to CD, so EF (is) to GH.
Thus, if four straight-lines are proportional, then similar, and similarly described, rectilinear figures (drawn) on them will also be proportional.
And if similar, and similarly described, rectilinear figures (drawn) on them are proportional then the straight-lines themselves will also be proportional.
(Which is) the very thing it was required to show.
Proposition 23
Equiangular parallelograms have to one another the ratio compounded out of (the ratios of) their sides.
Let AC and CF be equiangular parallelograms having angle BCD equal to ECG.
I say that parallelogram AC has to parallelogram CF the ratio compounded out of (the ratios of) their sides.
For let BC be laid down so as to be straight-on to CG.
Thus, DC is also straight-on to CE [Prop. 1.14].
And let the parallelogram DG have been completed.
And let some straight-line K have been laid down.
And let it be contrived that as BC (is) to CG, so K (is) to L, and as DC (is) to CE, so L (is) to M [Prop. 6.12].
Thus, the ratios of K to L and of L to M are the same as the ratios of the sides, (namely), BC to CG and DC to CE (respectively).
But, the ratio of K to M is compounded out of the ratio of K to L and (the ratio) of L to M.
Hence, K also has to M the ratio compounded out of (the ratios of) the sides (of the parallelograms).
And since as BC is to CG, so parallelogram AC (is) to CH [Prop. 6.1], but as BC (is) to CG, so K (is) to L, thus, also, as K (is) to L, so (parallelogram) AC (is) to CH.
Again, since as DC (is) to CE, so parallelogram CH (is) to CF [Prop. 6.1], but as DC (is) to CE, so L (is) to M, thus, also, as L (is) to M, so parallelogram CH (is) to parallelogram CF.
Therefore, since it was shown that as K (is) to L, so parallelogram AC (is) to parallelogram CH, and as L (is) to M, so parallelogram CH (is) to parallelogram CF, thus, via equality, as K is to M, so (parallelogram) AC (is) to parallelogram CF [Prop. 5.22].
And K has to M the ratio compounded out of (the ratios of) the sides (of the parallelograms).
Thus, (parallelogram) AC also has to (parallelogram) CF the ratio compounded out of (the ratio of) their sides.
Thus, equiangular parallelograms have to one another the ratio compounded out of (the ratio of) their sides.
(Which is) the very thing it was required to show.
Proposition 24
In any parallelogram the parallelograms about the diagonal are similar to the whole, and to one another.
Let ABCD be a parallelogram, and AC its diagonal.
And let EG and HK be parallelograms about AC.
I say that the parallelograms EG and HK are each similar to the whole (parallelogram) ABCD, and to one another.
For since EF has been drawn parallel to one of the sides BC of triangle ABC, proportionally, as BE is to EA, so CF (is) to FA [Prop. 6.2].
Again, since FG has been drawn parallel to one (of the sides) CD of triangle ACD, proportionally, as CF is to FA, so DG (is) to GA [Prop. 6.2].
But, as CF (is) to FA, so it was also shown (is) BE to EA.
And thus as BE (is) to EA, so DG (is) to GA.
And, thus, compounding, as BA (is) to AE, so DA (is) to AG [Prop. 5.18].
And, alternately, as BA (is) to AD, so EA (is) to AG [Prop. 5.16].
Thus, in parallelograms ABCD and EG the sides about the common angle BAD are proportional.
And since GF is parallel to DC, angle AFG is equal to DCA [Prop. 1.29].
And angle DAC (is) common to the two triangles ADC and AGF.
Thus, triangle ADC is equiangular to triangle AGF [Prop. 1.32].
So, for the same (reasons), triangle ACB is equiangular to triangle AFE, and the whole parallelogram ABCD is equiangular to parallelogram EG.
Thus, proportionally, as AD (is) to DC, so AG (is) to GF, and as DC (is) to CA, so GF (is) to FA, and as AC (is) to CB, so AF (is) to FE, and, further, as CB (is) to BA, so FE (is) to EA [Prop. 6.4].
And since it was shown that as DC is to CA, so GF (is) to FA, and as AC (is) to CB, so AF (is) to FE, thus, via equality, as DC is to CB, so GF (is) to FE [Prop. 5.22].
Thus, in parallelograms ABCD and EG the sides about the equal angles are proportional.
Thus, parallelogram ABCD is similar to parallelogram EG [Def. 6.1].
So, for the same (reasons), parallelogram ABCD is also similar to parallelogram KH.
Thus, parallelograms EG and HK are each similar to [parallelogram] ABCD.
And (rectilinear figures) similar to the same rectilinear figure are also similar to one another [Prop. 6.21].
Thus, parallelogram EG is also similar to parallelogram HK.
Thus, in any parallelogram the parallelograms about the diagonal are similar to the whole, and to one another.
(Which is) the very thing it was required to show.
Proposition 25
To construct a single (rectilinear figure) similar to a given rectilinear figure, and equal to a different given rectilinear figure.
Let ABC be the given rectilinear figure to which it is required to construct a similar (rectilinear figure), and D the (rectilinear figure) to which (the constructed figure) is required (to be) equal.
So it is required to construct a single (rectilinear figure) similar to ABC, and equal to D.
For let the parallelogram BE, equal to triangle ABC, have been applied to (the straight-line) BC [Prop. 1.44], and the parallelogram CM, equal to D, (have been applied) to (the straight-line) CE, in the angle FCE, which is equal to CBL [Prop. 1.45].
Thus, BC is straight-on to CF, and LE to EM [Prop. 1.14].
And let the mean proportion GH have been taken of BC and CF [Prop. 6.13].
And let KGH, similar, and similarly laid out, to ABC have been described on GH [Prop. 6.18].
And since as BC is to GH, so GH (is) to CF, and if three straight-lines are proportional then as the first is to the third, so the figure (described) on the first (is) to the similar, and similarly described, (figure) on the second [Prop. 6.19 corr.], thus as BC is to CF, so triangle ABC (is) to triangle KGH.
But, also, as BC (is) to CF, so parallelogram BE (is) to parallelogram EF [Prop. 6.1].
And, thus, as triangle ABC (is) to triangle KGH, so parallelogram BE (is) to parallelogram EF.
Thus, alternately, as triangle ABC (is) to parallelogram BE, so triangle KGH (is) to parallelogram EF [Prop. 5.16].
And triangle ABC (is) equal to parallelogram BE.
Thus, triangle KGH (is) also equal to parallelogram EF.
But, parallelogram EF is equal to D.
Thus, KGH is also equal to D.
And KGH is also similar to ABC.
Thus, a single (rectilinear figure) KGH has been constructed (which is) similar to the given rectilinear figure ABC, and equal to a different given (rectilinear figure) D.
(Which is) the very thing it was required to do.
Proposition 26
If from a parallelogram a(nother) parallelogram is subtracted (which is) similar, and similarly laid out, to the whole, having a common angle with it, then (the subtracted parallelogram) is about the same diagonal as the whole.
For, from parallelogram ABCD, let (parallelogram) AF have been subtracted (which is) similar, and similarly laid out, to ABCD, having the common angle DAB with it.
I say that ABCD is about the same diagonal as AF.
For (if) not, then, if possible, let AHC be [ ABCD 's] diagonal.
And producing GF, let it have been drawn through to (point) H (as GH).
And let HK have been drawn through (point) H, parallel to either of AD or BC [Prop. 1.31].
Therefore, since ABCD is about the same diagonal as KG, thus as DA is to AB, so GA (is) to AK [Prop. 6.24].
And, on account of the similarity of ABCD and EG, also, as DA (is) to AB, so GA (is) to AE.
Thus, also, as GA (is) to AK, so GA (is) to AE.
Thus, GA has the same ratio to each of AK and AE.
Thus, AE is equal to AK [Prop. 5.9], the lesser to the greater.
The very thing is impossible.
Thus, ABCD is not not about the same diagonal as AF.
Thus, parallelogram ABCD is about the same diagonal as parallelogram AF.
Thus, if from a parallelogram a(nother) parallelogram is subtracted (which is) similar, and similarly laid out, to the whole, having a common angle with it, then (the subtracted parallelogram) is about the same diagonal as the whole.
(Which is) the very thing it was required to show.
Proposition 27
Of all the parallelograms applied to the same straight-line, and falling short by parallelogrammic figures similar, and similarly laid out, to the (parallelogram) described on half (the straight-line), the greatest is the [parallelogram] applied to half (the straight-line) which (is) similar to (that parallelogram) by which it falls short.
Let AB be a straight-line, and let it have been cut in half at (point) C [Prop. 1.10].
And let the parallelogram AD have been applied to the straight-line AB, falling short by the parallelogrammic figure DB (which is) applied to half of AB ---that is to say, CB.
I say that of all the parallelograms applied to AB, and falling short by [parallelogrammic] figures similar, and similarly laid out, to DB, the greatest is AD.
For let the parallelogram AF have been applied to the straight-line AB, falling short by the parallelogrammic figure FB (which is) similar, and similarly laid out, to DB.
I say that AD is greater than AF.
For since parallelogram DB is similar to parallelogram FB, they are about the same diagonal [Prop. 6.26].
Let their (common) diagonal DB have been drawn, and let the (rest of the) figure have been described.
Therefore, since (complement) CF is equal to (complement) FE [Prop. 1.43], and (parallelogram) FB is common, the whole (parallelogram) CH is thus equal to the whole (parallelogram) KE.
But, (parallelogram) CH is equal to CG, since AC (is) also (equal) to CB [Prop. 6.1].
Thus, (parallelogram) GC is also equal to EK.
Let (parallelogram) CF have been added to both.
Thus, the whole (parallelogram) AF is equal to the gnomon LMN.
Hence, parallelogram DB ---that is to say, AD ---is greater than parallelogram AF.
Thus, for all parallelograms applied to the same straight-line, and falling short by a parallelogrammic figure similar, and similarly laid out, to the (parallelogram) described on half (the straight-line), the greatest is the [parallelogram] applied to half (the straight-line).
(Which is) the very thing it was required to show.
Proposition 28
To apply a parallelogram, equal to a given rectilinear figure, to a given straight-line, (the applied parallelogram) falling short by a parallelogrammic figure similar to a given (parallelogram).
It is necessary for the given rectilinear figure not to be greater than the (parallelogram) described on half (of the straight-line) and similar to the deficit.
Let AB be the given straight-line, and C the given rectilinear figure to which the (parallelogram) applied to AB is required (to be) equal, [being] not greater than the (parallelogram) described on half of AB and similar to the deficit, and D the (parallelogram) to which the deficit is required (to be) similar.
So it is required to apply a parallelogram, equal to the given rectilinear figure C, to the straight-line AB, falling short by a parallelogrammic figure which is similar to D.
Let AB have been cut in half at point E [Prop. 1.10], and let (parallelogram) EBFG, (which is) similar, and similarly laid out, to (parallelogram) D, have been described on EB [Prop. 6.18].
And let parallelogram AG have been completed.
Therefore, if AG is equal to C then the thing prescribed has happened.
For a parallelogram AG, equal to the given rectilinear figure C, has been applied to the given straight-line AB, falling short by a parallelogrammic figure GB which is similar to D.
And if not, let HE be greater than C.
And HE (is) equal to GB [Prop. 6.1].
Thus, GB (is) also greater than C.
So, let (parallelogram) KLMN have been constructed (so as to be) both similar, and similarly laid out, to D, and equal to the excess by which GB is greater than C [Prop. 6.25].
But, GB [is] similar to D.
Thus, KM is also similar to GB [Prop. 6.21].
Therefore, let KL correspond to GE, and LM to GF.
And since (parallelogram) GB is equal to (figure) C and (parallelogram) KM, GB is thus greater than KM.
Thus, GE is also greater than KL, and GF than LM.
Let GO be made equal to KL, and GP to LM [Prop. 1.3].
And let the parallelogram OGPQ have been completed.
Thus, GQ is equal and similar to KM.
Thus, GQ is also similar to GB [Prop. 6.21].
Thus, GQ and GB are about the same diagonal [Prop. 6.26].
Let GQB be their (common) diagonal, and let the (remainder of the) figure have been described.
Therefore, since BG is equal to C and KM, of which GQ is equal to KM, the remaining gnomon VWU is thus equal to the remainder C.
And since (the complement) PR is equal to (the complement) OS [Prop. 1.43], let (parallelogram) QB have been added to both.
Thus, the whole (parallelogram) PB is equal to the whole (parallelogram) OB.
But, OB is equal to TE, since side AE is equal to side EB [Prop. 6.1].
Thus, TE is also equal to PB.
Let (parallelogram) OS have been added to both.
Thus, the whole (parallelogram) TS is equal to the gnomon VWU.
But, gnomon VWU was shown (to be) equal to C.
Therefore, (parallelogram) TS is also equal to (figure) C.
Thus, the parallelogram ST, equal to the given rectilinear figure C, has been applied to the given straight-line AB, falling short by the parallelogrammic figure QB, which is similar to D.
(Which is) the very thing it was required to do.
Proposition 29
To apply a parallelogram, equal to a given rectilinear figure, to a given straight-line, (the applied parallelogram) overshooting by a parallelogrammic figure similar to a given (parallelogram).
Let AB be the given straight-line, and C the given rectilinear figure to which the (parallelogram) applied to AB is required (to be) equal, and D the (parallelogram) to which the excess is required (to be) similar.
So it is required to apply a parallelogram, equal to the given rectilinear figure C, to the given straight-line AB, overshooting by a parallelogrammic figure similar to D.
Let AB have been cut in half at (point) E [Prop. 1.10], and let the parallelogram BF, (which is) similar, and similarly laid out, to D, have been described on EB [Prop. 6.18].
And let (parallelogram) GH have been constructed (so as to be) both similar, and similarly laid out, to D, and equal to the sum of BF and C [Prop. 6.25].
And let KH correspond to FL, and KG to FE.
And since (parallelogram) GH is greater than (parallelogram) FB, KH is thus also greater than FL, and KG than FE.
Let FL and FE have been produced, and let FLM be (made) equal to KH, and FEN to KG [Prop. 1.3].
And let (parallelogram) MN have been completed.
Thus, MN is equal and similar to GH.
But, GH is similar to EL.
Thus, MN is also similar to EL [Prop. 6.21].
EL is thus about the same diagonal as MN [Prop. 6.26].
Let their (common) diagonal FO have been drawn, and let the (remainder of the) figure have been described.
And since (parallelogram) GH is equal to (parallelogram) EL and (figure) C, but GH is equal to (parallelogram) MN, MN is thus also equal to EL and C.
Let EL have been subtracted from both.
Thus, the remaining gnomon VWX is equal to (figure) C.
And since AE is equal to EB, (parallelogram) AN is also equal to (parallelogram) NB [Prop. 6.1], that is to say, (parallelogram) LP [Prop. 1.43].
Let (parallelogram) EO have been added to both.
Thus, the whole (parallelogram) AO is equal to the gnomon VWX.
But, the gnomon VWX is equal to (figure) C.
Thus, (parallelogram) AO is also equal to (figure) C.
Thus, the parallelogram AO, equal to the given rectilinear figure C, has been applied to the given straight-line AB, overshooting by the parallelogrammic figure QP which is similar to D, since PQ is also similar to EL [Prop. 6.24].
(Which is) the very thing it was required to do.
Proposition 30
To cut a given finite straight-line in extreme and mean ratio.
Let AB be the given finite straight-line.
So it is required to cut the straight-line AB in extreme and mean ratio.
Let the square BC have been described on AB [Prop. 1.46], and let the parallelogram CD, equal to BC, have been applied to AC, overshooting by the figure AD (which is) similar to BC [Prop. 6.29].
And BC is a square.
Thus, AD is also a square.
And since BC is equal to CD, let (rectangle) CE have been subtracted from both.
Thus, the remaining (rectangle) BF is equal to the remaining (square) AD.
And it is also equiangular to it.
Thus, the sides of BF and AD about the equal angles are reciprocally proportional [Prop. 6.14].
Thus, as FE is to ED, so AE (is) to EB.
And FE (is) equal to AB, and ED to AE.
Thus, as BA is to AE, so AE (is) to EB.
And AB (is) greater than AE.
Thus, AE (is) also greater than EB [Prop. 5.14].
Thus, the straight-line AB has been cut in extreme and mean ratio at E, and AE is its greater piece.
(Which is) the very thing it was required to do.
Proposition 31
In right-angled triangles, the figure (drawn) on the side subtending the right-angle is equal to the (sum of the) similar, and similarly described, figures on the sides surrounding the right-angle.
Let ABC be a right-angled triangle having the angle BAC a right-angle.
I say that the figure (drawn) on BC is equal to the (sum of the) similar, and similarly described, figures on BA and AC.
Let the perpendicular AD have been drawn [Prop. 1.12].
Therefore, since, in the right-angled triangle ABC, the (straight-line) AD has been drawn from the rightangle at A perpendicular to the base BC, the triangles ABD and ADC about the perpendicular are similar to the whole (triangle) ABC, and to one another [Prop. 6.8].
And since ABC is similar to ABD, thus as CB is to BA, so AB (is) to BD [Def. 6.1].
And since three straight-lines are proportional, as the first is to the third, so the figure (drawn) on the first is to the similar, and similarly described, (figure) on the second [Prop. 6.19 corr.].
Thus, as CB (is) to BD, so the figure (drawn) on CB (is) to the similar, and similarly described, (figure) on BA.
And so, for the same (reasons), as BC (is) to CD, so the figure (drawn) on BC (is) to the (figure) on CA.
Hence, also, as BC (is) to BD and DC, so the figure (drawn) on BC (is) to the (sum of the) similar, and similarly described, (figures) on BA and AC [Prop. 5.24].
And BC is equal to BD and DC.
Thus, the figure (drawn) on BC (is) also equal to the (sum of the) similar, and similarly described, figures on BA and AC [Prop. 5.9].
Thus, in right-angled triangles, the figure (drawn) on the side subtending the right-angle is equal to the (sum of the) similar, and similarly described, figures on the sides surrounding the right-angle.
(Which is) the very thing it was required to show.
Proposition 32
If two triangles, having two sides proportional to two sides, are placed together at a single angle such that the corresponding sides are also parallel, then the remaining sides of the triangles will be straight-on (with respect to one another).
Let ABC and DCE be two triangles having the two sides BA and AC proportional to the two sides DC and DE ---so that as AB (is) to AC, so DC (is) to DE ---and (having side) AB parallel to DC, and AC to DE.
I say that (side) BC is straight-on to CE.
For since AB is parallel to DC, and the straight-line AC has fallen across them, the alternate angles BAC and ACD are equal to one another [Prop. 1.29].
So, for the same (reasons), CDE is also equal to ACD.
And, hence, BAC is equal to CDE.
And since ABC and DCE are two triangles having the one angle at A equal to the one angle at D, and the sides about the equal angles proportional, (so that) as BA (is) to AC, so CD (is) to DE, triangle ABC is thus equiangular to triangle DCE [Prop. 6.6].
Thus, angle ABC is equal to DCE.
And (angle) ACD was also shown (to be) equal to BAC.
Thus, the whole (angle) ACE is equal to the two (angles) ABC and BAC.
Let ACB have been added to both.
Thus, ACE and ACB are equal to BAC, ACB, and CBA.
But, BAC, ABC, and ACB are equal to two right-angles [Prop. 1.32].
Thus, ACE and ACB are also equal to two right-angles.
Thus, the two straight-lines BC and CE, not lying on the same side, make adjacent angles ACE and ACB (whose sum is) equal to two right-angles with some straight-line AC, at the point C on it.
Thus, BC is straight-on to CE [Prop. 1.14].
Thus, if two triangles, having two sides proportional to two sides, are placed together at a single angle such that the corresponding sides are also parallel, then the remaining sides of the triangles will be straight-on (with respect to one another).
(Which is) the very thing it was required to show.
Proposition 33
In equal circles, angles have the same ratio as the (ratio of the) circumferences on which they stand, whether they are standing at the centers (of the circles) or at the circumferences.
Let ABC and DEF be equal circles, and let BGC and EHF be angles at their centers, G and H (respectively), and BAC and EDF (angles) at their circumferences.
I say that as circumference BC is to circumference EF, so angle BGC (is) to EHF, and (angle) BAC to EDF.
For let any number whatsoever of consecutive (circumferences), CK and KL, be made equal to circumference BC, and any number whatsoever, FM and MN, to circumference EF.
And let GK, GL, HM, and HN have been joined.
Therefore, since circumferences BC, CK, and KL are equal to one another, angles BGC, CGK, and KGL are also equal to one another [Prop. 3.27].
Thus, as many times as circumference BL is (divisible) by BC, so many times is angle BGL also (divisible) by BGC.
And so, for the same (reasons), as many times as circumference EN is (divisible) by EF, so many times is angle NHE also (divisible) by EHF.
Thus, if circumference BL is equal to circumference EN then angle BGL is also equal to EHN [Prop. 3.27], and if circumference BL is greater than circumference EN then angle BGL is also greater than EHN, and if ( BL is) less (than EN then BGL is also) less (than EHN).
So there are four magnitudes, two circumferences BC and EF, and two angles BGC and EHF.
And equal multiples have been taken of circumference BC and angle BGC, (namely) circumference BL and angle BGL, and of circumference EF and angle EHF, (namely) circumference EN and angle EHN.
And it has been shown that if circumference BL exceeds circumference EN then angle BGL also exceeds angle EHN, and if ( BL is) equal (to EN then BGL is also) equal (to EHN), and if ( BL is) less (than EN then BGL is also) less (than EHN).
Thus, as circumference BC (is) to EF, so angle BGC (is) to EHF [Def. 5.5].
But as angle BGC (is) to EHF, so (angle) BAC (is) to EDF [Prop. 5.15].
For the former (are) double the latter (respectively) [Prop. 3.20].
Thus, also, as circumference BC (is) to circumference EF, so angle BGC (is) to EHF, and BAC to EDF.
Thus, in equal circles, angles have the same ratio as the (ratio of the) circumferences on which they stand, whether they are standing at the centers (of the circles) or at the circumferences.
(Which is) the very thing it was required to show.