Book 4 — Construction of Rectilinear Figures In and Around Circles ◧
Definitions
1. A rectilinear figure is said to be inscribed in a(nother) rectilinear figure when the respective angles of the inscribed figure touch the respective sides of the (figure) in which it is inscribed.
2. And, similarly, a (rectilinear) figure is said to be circumscribed about a(nother rectilinear) figure when the respective sides of the circumscribed (figure) touch the respective angles of the (figure) about which it is circumscribed.
3. A rectilinear figure is said to be inscribed in a circle when each angle of the inscribed (figure) touches the circumference of the circle.
4. And a rectilinear figure is said to be circumscribed about a circle when each side of the circumscribed (figure) touches the circumference of the circle.
5. And, similarly, a circle is said to be inscribed in a (rectilinear) figure when the circumference of the circle touches each side of the (figure) in which it is inscribed.
6. And a circle is said to be circumscribed about a rectilinear (figure) when the circumference of the circle touches each angle of the (figure) about which it is circumscribed.
7. A straight-line is said to be inserted into a circle when its extemities are on the circumference of the circle.
Proposition 1
To insert a straight-line equal to a given straight-line into a circle, (the latter straight-line) not being greater than the diameter of the circle.
Let ABC be the given circle, and D the given straight-line (which is) not greater than the diameter of the circle.
So it is required to insert a straight-line, equal to the straight-line D, into the circle ABC.
Let a diameter BC of circle ABC have been drawn.
Therefore, if BC is equal to D then that (which) was prescribed has taken place.
For the (straight-line) BC, equal to the straight-line D, has been inserted into the circle ABC.
And if BC is greater than D then let CE be made equal to D [Prop. 1.3], and let the circle EAF have been drawn with center C and radius CE.
And let CA have been joined.
Therefore, since the point C is the center of circle EAF, CA is equal to CE.
But, CE is equal to D.
Thus, D is also equal to CA.
Thus, CA, equal to the given straight-line D, has been inserted into the given circle ABC.
(Which is) the very thing it was required to do.
Proposition 2
To inscribe a triangle, equiangular with a given triangle, in a given circle.
Let ABC be the given circle, and DEF the given triangle.
So it is required to inscribe a triangle, equiangular with triangle DEF, in circle ABC.
Let GH have been drawn touching circle ABC at A.
And let (angle) HAC, equal to angle DEF, have been constructed on the straight-line AH at the point A on it, and (angle) GAB, equal to [angle] DFE, on the straight-line AG at the point A on it [Prop. 1.23].
And let BC have been joined.
Therefore, since some straight-line AH touches the circle ABC, and the straight-line AC has been drawn across (the circle) from the point of contact A, (angle) HAC is thus equal to the angle ABC in the alternate segment of the circle [Prop. 3.32].
But, HAC is equal to DEF.
Thus, angle ABC is also equal to DEF.
So, for the same (reasons), ACB is also equal to DFE.
Thus, the remaining (angle) BAC is equal to the remaining (angle) EDF [Prop. 1.32].
Thus, a triangle, equiangular with the given triangle, has been inscribed in the given circle.
(Which is) the very thing it was required to do.
Proposition 3
To circumscribe a triangle, equiangular with a given triangle, about a given circle.
Let ABC be the given circle, and DEF the given triangle.
So it is required to circumscribe a triangle, equiangular with triangle DEF, about circle ABC.
Let EF have been produced in each direction to points G and H.
And let the center K of circle ABC have been found [Prop. 3.1].
And let the straight-line KB have been drawn, at random, across ( ABC).
And let (angle) BKA, equal to angle DEG, have been constructed on the straight-line KB at the point K on it, and (angle) BKC, equal to DFH [Prop. 1.23].
And let the (straight-lines) LAM, MBN, and NCL have been drawn through the points A, B, and C (respectively), touching the circle ABC.
And since LM, MN, and NL touch circle ABC at points A, B, and C (respectively), and KA, KB, and KC are joined from the center K to points A, B, and C (respectively), the angles at points A, B, and C are thus right-angles [Prop. 3.18].
And since the (sum of the) four angles of quadrilateral AMBK is equal to four right-angles, inasmuch as AMBK (can) also (be) divided into two triangles [Prop. 1.32], and angles KAM and KBM are (both) right-angles, the (sum of the) remaining (angles), AKB and AMB, is thus equal to two right-angles.
And DEG and DEF is also equal to two right-angles [Prop. 1.13].
Thus, AKB and AMB is equal to DEG and DEF, of which AKB is equal to DEG.
Thus, the remainder AMB is equal to the remainder DEF.
So, similarly, it can be shown that LNB is also equal to DFE.
Thus, the remaining (angle) MLN is also equal to the [remaining] (angle) EDF [Prop. 1.32].
Thus, triangle LMN is equiangular with triangle DEF.
And it has been drawn around circle ABC.
Thus, a triangle, equiangular with the given triangle, has been circumscribed about the given circle.
(Which is) the very thing it was required to do.
Proposition 4
To inscribe a circle in a given triangle.
Let ABC be the given triangle.
So it is required to inscribe a circle in triangle ABC.
Let the angles ABC and ACB have been cut in half by the straight-lines BD and CD (respectively) [Prop. 1.9], and let them meet one another at point D, and let DE, DF, and DG have been drawn from point D, perpendicular to the straight-lines AB, BC, and CA (respectively) [Prop. 1.12].
And since angle ABD is equal to CBD, and the right-angle BED is also equal to the right-angle BFD, EBD and FBD are thus two triangles having two angles equal to two angles, and one side equal to one side---the (one) subtending one of the equal angles (which is) common to the (triangles)---(namely), BD.
Thus, they will also have the remaining sides equal to the (corresponding) remaining sides [Prop. 1.26].
Thus, DE (is) equal to DF.
So, for the same (reasons), DG is also equal to DF.
Thus, the three straight-lines DE, DF, and DG are equal to one another.
Thus, the circle drawn with center D, and radius one of E, F, or G, will also go through the remaining points, and will touch the straight-lines AB, BC, and CA, on account of the angles at E, F, and G being right-angles.
For if it cuts (one of) them then it will be a (straight-line) drawn at right-angles to a diameter of the circle, from its extremity, falling inside the circle.
The very thing was shown (to be) absurd [Prop. 3.16].
Thus, the circle drawn with center D, and radius one of E, F, or G, does not cut the straight-lines AB, BC, and CA.
Thus, it will touch them and will be the circle inscribed in triangle ABC.
Let it have been (so) inscribed, like FGE (in the figure).
Thus, the circle EFG has been inscribed in the given triangle ABC.
(Which is) the very thing it was required to do.
Proposition 5
To circumscribe a circle about a given triangle.
Let ABC be the given triangle.
So it is required to circumscribe a circle about the given triangle ABC.
Let the straight-lines AB and AC have been cut in half at points D and E (respectively) [Prop. 1.10].
And let DF and EF have been drawn from points D and E, at right-angles to AB and AC (respectively) [Prop. 1.11].
So ( DF and EF) will surely either meet inside triangle ABC, on the straight-line BC, or beyond BC.
Let them, first of all, meet inside (triangle ABC) at (point) F, and let FB, FC, and FA have been joined.
And since AD is equal to DB, and DF is common and at right-angles, the base AF is thus equal to the base FB [Prop. 1.4].
So, similarly, we can show that CF is also equal to AF.
So that FB is also equal to FC.
Thus, the three (straight-lines) FA, FB, and FC are equal to one another.
Thus, the circle drawn with center F, and radius one of A, B, or C, will also go through the remaining points.
And the circle will have been circumscribed about triangle ABC.
Let it have been (so) circumscribed, like ABC (in the first diagram from the left).
And so, let DF and EF meet on the straight-line BC at (point) F, like in the second diagram (from the left).
And let AF have been joined.
So, similarly, we can show that point F is the center of the circle circumscribed about triangle ABC.
And so, let DF and EF meet outside triangle ABC, again at (point) F, like in the third diagram (from the left).
And let AF, BF, and CF have been joined.
And, again, since AD is equal to DB, and DF is common and at right-angles, the base AF is thus equal to the base BF [Prop. 1.4].
So, similarly, we can show that CF is also equal to AF.
So that BF is also equal to FC.
Thus, [again] the circle drawn with center F, and radius one of FA, FB, and FC, will also go through the remaining points.
And it will have been circumscribed about triangle ABC.
Thus, a circle has been circumscribed about the given triangle.
(Which is) the very thing it was required to do.
Proposition 6
To inscribe a square in a given circle.
Let ABCD be the given circle.
So it is required to inscribe a square in circle ABCD.
Let two diameters of circle ABCD, AC and BD, have been drawn at right-angles to one another.
And let AB, BC, CD, and DA have been joined.
And since BE is equal to ED, for E (is) the center (of the circle), and EA is common and at right-angles, the base AB is thus equal to the base AD [Prop. 1.4].
So, for the same (reasons), each of BC and CD is equal to each of AB and AD.
Thus, the quadrilateral ABCD is equilateral.
So I say that (it is) also right-angled.
For since the straight-line BD is a diameter of circle ABCD, BAD is thus a semi-circle.
Thus, angle BAD (is) a right-angle [Prop. 3.31].
So, for the same (reasons), (angles) ABC, BCD, and CDA are also each right-angles.
Thus, the quadrilateral ABCD is right-angled.
And it was also shown (to be) equilateral.
Thus, it is a square [Def. 1.22].
And it has been inscribed in circle ABCD.
Thus, the square ABCD has been inscribed in the given circle.
(Which is) the very thing it was required to do.
Proposition 7
To circumscribe a square about a given circle.
Let ABCD be the given circle.
So it is required to circumscribe a square about circle ABCD.
Let two diameters of circle ABCD, AC and BD, have been drawn at right-angles to one another.
And let FG, GH, HK, and KF have been drawn through points A, B, C, and D (respectively), touching circle ABCD.
Therefore, since FG touches circle ABCD, and EA has been joined from the center E to the point of contact A, the angles at A are thus right-angles [Prop. 3.18].
So, for the same (reasons), the angles at points B, C, and D are also right-angles.
And since angle AEB is a right-angle, and EBG is also a right-angle, GH is thus parallel to AC [Prop. 1.29].
So, for the same (reasons), AC is also parallel to FK.
So that GH is also parallel to FK [Prop. 1.30].
So, similarly, we can show that GF and HK are each parallel to BED.
Thus, GK, GC, AK, FB, and BK are (all) parallelograms.
Thus, GF is equal to HK, and GH to FK [Prop. 1.34].
And since AC is equal to BD, but AC (is) also (equal) to each of GH and FK, and BD is equal to each of GF and HK [Prop. 1.34], the quadrilateral FGHK is thus equilateral.
So I say that (it is) also right-angled.
For since GBEA is a parallelogram, and AEB is a right-angle, AGB is thus also a right-angle [Prop. 1.34].
So, similarly, we can show that the angles at H, K, and F are also right-angles.
Thus, FGHK is right-angled.
And it was also shown (to be) equilateral.
Thus, it is a square [Def. 1.22].
And it has been circumscribed about circle ABCD.
Thus, a square has been circumscribed about the given circle.
(Which is) the very thing it was required to do.
Proposition 8
To inscribe a circle in a given square.
Let the given square be ABCD.
So it is required to inscribe a circle in square ABCD.
Let AD and AB each have been cut in half at points E and F (respectively) [Prop. 1.10].
And let EH have been drawn through E, parallel to either of AB or CD, and let FK have been drawn through F, parallel to either of AD or BC [Prop. 1.31].
Thus, AK, KB, AH, HD, AG, GC, BG, and GD are each parallelograms, and their opposite sides [are] manifestly equal [Prop. 1.34].
And since AD is equal to AB, and AE is half of AD, and AF half of AB, AE (is) thus also equal to AF.
So that the opposite (sides are) also (equal).
Thus, FG (is) also equal to GE.
So, similarly, we can also show that each of GH and GK is equal to each of FG and GE.
Thus, the four (straight-lines) GE, GF, GH, and GK [are] equal to one another.
Thus, the circle drawn with center G, and radius one of E, F, H, or K, will also go through the remaining points.
And it will touch the straight-lines AB, BC, CD, and DA, on account of the angles at E, F, H, and K being right-angles.
For if the circle cuts AB, BC, CD, or DA, then a (straight-line) drawn at right-angles to a diameter of the circle, from its extremity, will fall inside the circle.
The very thing was shown (to be) absurd [Prop. 3.16].
Thus, the circle drawn with center G, and radius one of E, F, H, or K, does not cut the straight-lines AB, BC, CD, or DA.
Thus, it will touch them, and will have been inscribed in the square ABCD.
Thus, a circle has been inscribed in the given square.
(Which is) the very thing it was required to do.
Proposition 9
To circumscribe a circle about a given square.
Let ABCD be the given square.
So it is required to circumscribe a circle about square ABCD.
AC and BD being joined, let them cut one another at E.
And since DA is equal to AB, and AC (is) common, the two (straight-lines) DA, AC are thus equal to the two (straight-lines) BA, AC.
And the base DC (is) equal to the base BC.
Thus, angle DAC is equal to angle BAC [Prop. 1.8].
Thus, the angle DAB has been cut in half by AC.
So, similarly, we can show that ABC, BCD, and CDA have each been cut in half by the straight-lines AC and DB.
And since angle DAB is equal to ABC, and EAB is half of DAB, and EBA half of ABC, EAB is thus also equal to EBA.
So that side EA is also equal to EB [Prop. 1.6].
So, similarly, we can show that each of the [straight-lines] EA and EB are also equal to each of EC and ED.
Thus, the four (straight-lines) EA, EB, EC, and ED are equal to one another.
Thus, the circle drawn with center E, and radius one of A, B, C, or D, will also go through the remaining points, and will have been circumscribed about the square ABCD.
Let it have been (so) circumscribed, like ABCD (in the figure).
Thus, a circle has been circumscribed about the given square.
(Which is) the very thing it was required to do.
Proposition 10
To construct an isosceles triangle having each of the angles at the base double the remaining (angle).
Let some straight-line AB be taken, and let it have been cut at point C so that the rectangle contained by AB and BC is equal to the square on CA [Prop. 2.11].
And let the circle BDE have been drawn with center A, and radius AB.
And let the straight-line BD, equal to the straight-line AC, being not greater than the diameter of circle BDE, have been inserted into circle BDE [Prop. 4.1].
And let AD and DC have been joined.
And let the circle ACD have been circumscribed about triangle ACD [Prop. 4.5].
And since the (rectangle contained) by AB and BC is equal to the (square) on AC, and AC (is) equal to BD, the (rectangle contained) by AB and BC is thus equal to the (square) on BD.
And since some point B has been taken outside of circle ACD, and two straight-lines BA and BD have radiated from B towards the circle ACD, and (one) of them cuts (the circle), and (the other) meets (the circle), and the (rectangle contained) by AB and BC is equal to the (square) on BD, BD thus touches circle ACD [Prop. 3.37].
Therefore, since BD touches (the circle), and DC has been drawn across (the circle) from the point of contact D, the angle BDC is thus equal to the angle DAC in the alternate segment of the circle [Prop. 3.32].
Therefore, since BDC is equal to DAC, let CDA have been added to both.
Thus, the whole of BDA is equal to the two (angles) CDA and DAC.
But, the external (angle) BCD is equal to CDA and DAC [Prop. 1.32].
Thus, BDA is also equal to BCD.
But, BDA is equal to CBD, since the side AD is also equal to AB [Prop. 1.5].
So that DBA is also equal to BCD.
Thus, the three (angles) BDA, DBA, and BCD are equal to one another.
And since angle DBC is equal to BCD, side BD is also equal to side DC [Prop. 1.6].
But, BD was assumed (to be) equal to CA.
Thus, CA is also equal to CD.
So that angle CDA is also equal to angle DAC [Prop. 1.5].
Thus, CDA and DAC is double DAC.
But BCD (is) equal to CDA and DAC.
Thus, BCD is also double CAD.
And BCD (is) equal to to each of BDA and DBA.
Thus, BDA and DBA are each double DAB.
Thus, the isosceles triangle ABD has been constructed having each of the angles at the base BD double the remaining (angle).
(Which is) the very thing it was required to do.
Proposition 11
To inscribe an equilateral and equiangular pentagon in a given circle.
Let ABCDE be the given circle.
So it is required to inscribed an equilateral and equiangular pentagon in circle ABCDE.
Let the the isosceles triangle FGH be set up having each of the angles at G and H double the (angle) at F [Prop. 4.10].
And let triangle ACD, equiangular to FGH, have been inscribed in circle ABCDE, such that CAD is equal to the angle at F, and the (angles) at G and H (are) equal to ACD and CDA, respectively [Prop. 4.2].
Thus, ACD and CDA are each double CAD.
So let ACD and CDA have been cut in half by the straight-lines CE and DB, respectively [Prop. 1.9].
And let AB, BC, DE and EA have been joined.
Therefore, since angles ACD and CDA are each double CAD, and are cut in half by the straight-lines CE and DB, the five angles DAC, ACE, ECD, CDB, and BDA are thus equal to one another.
And equal angles stand upon equal circumferences [Prop. 3.26].
Thus, the five circumferences AB, BC, CD, DE, and EA are equal to one another [Prop. 3.29].
Thus, the pentagon ABCDE is equilateral.
So I say that (it is) also equiangular.
For since the circumference AB is equal to the circumference DE, let BCD have been added to both.
Thus, the whole circumference ABCD is equal to the whole circumference EDCB.
And the angle AED stands upon circumference ABCD, and angle BAE upon circumference EDCB.
Thus, angle BAE is also equal to AED [Prop. 3.27].
So, for the same (reasons), each of the angles ABC, BCD, and CDE is also equal to each of BAE and AED.
Thus, pentagon ABCDE is equiangular.
And it was also shown (to be) equilateral.
Thus, an equilateral and equiangular pentagon has been inscribed in the given circle.
(Which is) the very thing it was required to do.
Proposition 12
To circumscribe an equilateral and equiangular pentagon about a given circle.
Let ABCDE be the given circle.
So it is required to circumscribe an equilateral and equiangular pentagon about circle ABCDE.
Let A, B, C, D, and E have been conceived as the angular points of a pentagon having been inscribed (in circle ABCDE) [Prop. 3.11], such that the circumferences AB, BC, CD, DE, and EA are equal.
And let GH, HK, KL, LM, and MG have been drawn through (points) A, B, C, D, and E (respectively), touching the circle.
And let the center F of the circle ABCDE have been found [Prop. 3.1].
And let FB, FK, FC, FL, and FD have been joined.
And since the straight-line KL touches (circle) ABCDE at C, and FC has been joined from the center F to the point of contact C, FC is thus perpendicular to KL [Prop. 3.18].
Thus, each of the angles at C is a right-angle.
So, for the same (reasons), the angles at B and D are also right-angles.
And since angle FCK is a right-angle, the (square) on FK is thus equal to the (sum of the squares) on FC and CK [Prop. 1.47].
So, for the same (reasons), the (square) on FK is also equal to the (sum of the squares) on FB and BK.
So that the (sum of the squares) on FC and CK is equal to the (sum of the squares) on FB and BK, of which the (square) on FC is equal to the (square) on FB.
Thus, the remaining (square) on CK is equal to the remaining (square) on BK.
Thus, BK (is) equal to CK.
And since FB is equal to FC, and FK (is) common, the two (straight-lines) BF, FK are equal to the two (straight-lines) CF, FK.
And the base BK [is] equal to the base CK.
Thus, angle BFK is equal to [angle] KFC [Prop. 1.8].
And BKF (is equal) to FKC [Prop. 1.8].
Thus, BFC (is) double KFC, and BKC (is double) FKC.
So, for the same (reasons), CFD is also double CFL, and DLC (is also double) FLC.
And since circumference BC is equal to CD, angle BFC is also equal to CFD [Prop. 3.27].
And BFC is double KFC, and DFC (is double) LFC.
Thus, KFC is also equal to LFC.
And angle FCK is also equal to FCL.
So, FKC and FLC are two triangles having two angles equal to two angles, and one side equal to one side, (namely) their common (side) FC.
Thus, they will also have the remaining sides equal to the (corresponding) remaining sides, and the remaining angle to the remaining angle [Prop. 1.26].
Thus, the straight-line KC (is) equal to CL, and the angle FKC to FLC.
And since KC is equal to CL, KL (is) thus double KC.
So, for the same (reasons), it can be shown that HK (is) also double BK.
And BK is equal to KC.
Thus, HK is also equal to KL.
So, similarly, each of HG, GM, and ML can also be shown (to be) equal to each of HK and KL.
Thus, pentagon GHKLM is equilateral.
So I say that (it is) also equiangular.
For since angle FKC is equal to FLC, and HKL was shown (to be) double FKC, and KLM double FLC, HKL is thus also equal to KLM.
So, similarly, each of KHG, HGM, and GML can also be shown (to be) equal to each of HKL and KLM.
Thus, the five angles GHK, HKL, KLM, LMG, and MGH are equal to one another.
Thus, the pentagon GHKLM is equiangular.
And it was also shown (to be) equilateral, and has been circumscribed about circle ABCDE.
(Which is) the very thing it was required to do.
Proposition 13
To inscribe a circle in a given pentagon, which is equilateral and equiangular.
Let ABCDE be the given equilateral and equiangular pentagon.
So it is required to inscribe a circle in pentagon ABCDE.
For let angles BCD and CDE have each been cut in half by each of the straight-lines CF and DF (respectively) [Prop. 1.9].
And from the point F, at which the straight-lines CF and DF meet one another, let the straight-lines FB, FA, and FE have been joined.
And since BC is equal to CD, and CF (is) common, the two (straight-lines) BC, CF are equal to the two (straight-lines) DC, CF.
And angle BCF [is] equal to angle DCF.
Thus, the base BF is equal to the base DF, and triangle BCF is equal to triangle DCF, and the remaining angles will be equal to the (corresponding) remaining angles which the equal sides subtend [Prop. 1.4].
Thus, angle CBF (is) equal to CDF.
And since CDE is double CDF, and CDE (is) equal to ABC, and CDF to CBF, CBA is thus also double CBF.
Thus, angle ABF is equal to FBC.
Thus, angle ABC has been cut in half by the straight-line BF.
So, similarly, it can be shown that BAE and AED have been cut in half by the straight-lines FA and FE, respectively.
So let FG, FH, FK, FL, and FM have been drawn from point F, perpendicular to the straight-lines AB, BC, CD, DE, and EA (respectively) [Prop. 1.12].
And since angle HCF is equal to KCF, and the right-angle FHC is also equal to the [right-angle] FKC, FHC and FKC are two triangles having two angles equal to two angles, and one side equal to one side, (namely) their common (side) FC, subtending one of the equal angles.
Thus, they will also have the remaining sides equal to the (corresponding) remaining sides [Prop. 1.26].
Thus, the perpendicular FH (is) equal to the perpendicular FK.
So, similarly, it can be shown that FL, FM, and FG are each equal to each of FH and FK.
Thus, the five straight-lines FG, FH, FK, FL, and FM are equal to one another.
Thus, the circle drawn with center F, and radius one of G, H, K, L, or M, will also go through the remaining points, and will touch the straight-lines AB, BC, CD, DE, and EA, on account of the angles at points G, H, K, L, and M being right-angles.
For if it does not touch them, but cuts them, it follows that a (straight-line) drawn at right-angles to the diameter of the circle, from its extremity, falls inside the circle.
The very thing was shown (to be) absurd [Prop. 3.16].
Thus, the circle drawn with center F, and radius one of G, H, K, L, or M, does not cut the straight-lines AB, BC, CD, DE, or EA.
Thus, it will touch them.
Let it have been drawn, like GHKLM (in the figure).
Thus, a circle has been inscribed in the given pentagon which is equilateral and equiangular.
(Which is) the very thing it was required to do.
Proposition 14
To circumscribe a circle about a given pentagon which is equilateral and equiangular.
Let ABCDE be the given pentagon which is equilateral and equiangular.
So it is required to circumscribe a circle about the pentagon ABCDE.
So let angles BCD and CDE have been cut in half by the (straight-lines) CF and DF, respectively [Prop. 1.9].
And let the straight-lines FB, FA, and FE have been joined from point F, at which the straight-lines meet, to the points B, A, and E (respectively).
So, similarly, to the (proposition) before this (one), it can be shown that angles CBA, BAE, and AED have also been cut in half by the straight-lines FB, FA, and FE, respectively.
And since angle BCD is equal to CDE, and FCD is half of BCD, and CDF half of CDE, FCD is thus also equal to FDC.
So that side FC is also equal to side FD [Prop. 1.6].
So, similarly, it can be shown that FB, FA, and FE are also each equal to each of FC and FD.
Thus, the five straight-lines FA, FB, FC, FD, and FE are equal to one another.
Thus, the circle drawn with center F, and radius one of FA, FB, FC, FD, or FE, will also go through the remaining points, and will have been circumscribed.
Let it have been (so) circumscribed, and let it be ABCDE.
Thus, a circle has been circumscribed about the given pentagon, which is equilateral and equiangular.
(Which is) the very thing it was required to do.
Proposition 15
To inscribe an equilateral and equiangular hexagon in a given circle.
Let ABCDEF be the given circle.
So it is required to inscribe an equilateral and equiangular hexagon in circle ABCDEF.
Let the diameter AD of circle ABCDEF have been drawn, and let the center G of the circle have been found [Prop. 3.1].
And let the circle EGCH have been drawn, with center D, and radius DG.
And EG and CG being joined, let them have been drawn across (the circle) to points B and F (respectively).
And let AB, BC, CD, DE, EF, and FA have been joined.
I say that the hexagon ABCDEF is equilateral and equiangular.
For since point G is the center of circle ABCDEF, GE is equal to GD.
Again, since point D is the center of circle GCH, DE is equal to DG.
But, GE was shown (to be) equal to GD.
Thus, GE is also equal to ED.
Thus, triangle EGD is equilateral.
Thus, its three angles EGD, GDE, and DEG are also equal to one another, inasmuch as the angles at the base of isosceles triangles are equal to one another [Prop. 1.5].
And the three angles of the triangle are equal to two right-angles [Prop. 1.32].
Thus, angle EGD is one third of two right-angles.
So, similarly, DGC can also be shown (to be) one third of two right-angles.
And since the straight-line CG, standing on EB, makes adjacent angles EGC and CGB equal to two right-angles [Prop. 1.13], the remaining angle CGB is thus also one third of two right-angles.
Thus, angles EGD, DGC, and CGB are equal to one another.
And hence the (angles) opposite to them BGA, AGF, and FGE are also equal [Prop. 1.15].
Thus, the six angles EGD, DGC, CGB, BGA, AGF, and FGE are equal to one another.
And equal angles stand on equal circumferences [Prop. 3.26].
Thus, the six circumferences AB, BC, CD, DE, EF, and FA are equal to one another.
And equal circumferences are subtended by equal straight-lines [Prop. 3.29].
Thus, the six straight-lines ( AB, BC, CD, DE, EF, and FA) are equal to one another.
Thus, hexagon ABCDEF is equilateral.
So, I say that (it is) also equiangular.
For since circumference FA is equal to circumference ED, let circumference ABCD have been added to both.
Thus, the whole of FABCD is equal to the whole of EDCBA.
And angle FED stands on circumference FABCD, and angle AFE on circumference EDCBA.
Thus, angle AFE is equal to DEF [Prop. 3.27].
Similarly, it can also be shown that the remaining angles of hexagon ABCDEF are individually equal to each of the angles AFE and FED.
Thus, hexagon ABCDEF is equiangular.
And it was also shown (to be) equilateral.
And it has been inscribed in circle ABCDE.
Thus, an equilateral and equiangular hexagon has been inscribed in the given circle.
(Which is) the very thing it was required to do.
Corollary
So, from this, (it is) manifest that a side of the hexagon is equal to the radius of the circle.
And similarly to a pentagon, if we draw tangents to the circle through the (sixfold) divisions of the (circumference of the) circle, an equilateral and equiangular hexagon can be circumscribed about the circle, analogously to the aforementioned pentagon.
And, further, by (means) similar to the aforementioned pentagon, we can inscribe and circumscribe a circle in (and about) a given hexagon.
(Which is) the very thing it was required to do.
Proposition 16
To inscribe an equilateral and equiangular fifteen-sided figure in a given circle.
Let ABCD be the given circle.
So it is required to inscribe an equilateral and equiangular fifteen-sided figure in circle ABCD.
Let the side AC of an equilateral triangle inscribed in (the circle) [Prop. 4.2], and (the side) AB of an (inscribed) equilateral pentagon [Prop. 4.11], have been inscribed in circle ABCD.
Thus, just as the circle ABCD is (made up) of fifteen equal pieces, the circumference ABC, being a third of the circle, will be (made up) of five such (pieces), and the circumference AB, being a fifth of the circle, will be (made up) of three.
Thus, the remainder BC (will be made up) of two equal (pieces).
Let (circumference) BC have been cut in half at E [Prop. 3.30].
Thus, each of the circumferences BE and EC is one fifteenth of the circle ABCDE.
Thus, if, joining BE and EC, we continuously insert straight-lines equal to them into circle ABCD [Prop. 4.1], then an equilateral and equiangular fifteen-sided figure will have been inserted into (the circle).
(Which is) the very thing it was required to do.
And similarly to the pentagon, if we draw tangents to the circle through the (fifteenfold) divisions of the (circumference of the) circle, we can circumscribe an equilateral and equiangular fifteen-sided figure about the circle.
And, further, through similar proofs to the pentagon, we can also inscribe and circumscribe a circle in (and about) a given fifteen-sided figure.