Book IV.

a 10. I.

b II. I. C 4. I.

d Def. 6.

It is required to describe a circle about the triangle ABC. Bifect the fides AB, AC, in the points D, E; from which draw DF, FE, at right angles to AB, AC, which will meet one another, either within the triangle, upon one of the fides BC, or without the triangle, in the point F.

In either cafe, because AD is equal to DB, and DF common, the two fides AD, DF, are equal to BD, DF; and the angle BDF to ADF; therefore the base BF is equal to the bafe AF. For the fame reafon AF is equal to FC: Therefore, if, with the center F, and either of the distances AF, FB, or FC, a circle is defcribed, it will touch all the angles of the triangled: Which was required.

PRO P. VI. PRO B.

a II. I. 4. I.

C 31. 3.

It is required to infcribe a fquare in the given circle ABC. Draw the diameters BD and AC at right angles to, and bifecting each other in E; join BA, AD, DC, CB; then ADCB is a fquare.

For the two fides BE, EA, are equal to DE, EA, and the' angle BEA to AED; therefore the bafe BA is equal to AD b. For the fame reafon AD is equal to DC, and DC to CB; therefore the four fides are equal: But the angle BAD is a right angle; therefore ADC, DCB, ABC, are each right angles; therefore the figure is a fquared: Which was requi

d def. 30. 1. red.

a II. I.

b 16. 3.

C 29. I, d 30. I.

It is required to defcribe a fquare about the circle ABCD. Draw the two diameters AC, BD, of the circle, cutting each other at right angles "; and through the points A, B, C, D, draw FG, GH, HK, KF, tangents to the circle ABCD.

For, because the angles GAE, AEB, are right angles, GF is parallel to BD. For the fame reafon HK is parallel to BD; therefore GF is parallel to HK. For the fame reason, GH is

f Ax. I. I.

parallel to FK; therefore GHKF is a parallelogram; but GF, Book IV, BD, are equal, and likewife BD equal to HK; therefore GF is equal to HK f. For the fame reason, GH is equal to FK; fore 34. 1. each is equal to AC; but AC is equal to BD; therefore GH is equal to GF; therefore the four fides are equal; but the angles at G, F, are right ones; for each is equal to GAE, or FAE ;b 16. 39 therefore all the 'angles are right ones; therefore GHKF is a fquare, defcribed about the circle ABCD: Which was to be done.

b

It is required to inscribe a circle in the fquare ABCD.

Bifect the fides AB, AD, in the points E, F, through Ea 10. 1. draw EH parallel to AB, and through F draw FK parallel tob 31. 1. AD or BC; then AE, FG, are equal to one another, and € 34. I. likewife ED, GK; but AE is equal to ED; therefore FG is equal to GK; but AF is equal to AE ; therefore EG is equal d Ax. 7. 1, to FG, and FB to GH; therefore GE, GF, GH, GK, are equal; therefore if, with the center G, and either of these diftances, a circle is defcribed, it will touch the fquare in the points E, F, H, K; for if not, let it cut the fquare, then a right line drawn at right angles to the diameter of a circle will fall within the circle; which cannot be ; therefore the circlee 16. 3. EFHK is infcribed in the fquare ABCD: Which was required.

PRO P. IX. PRO B.

Ꭲ

O defcribe a circle about a given fquare.

It is required to defcribe a circle about the given fquare ABCD.

Join AC, BD, mutually cutting each other in the point E. For, because AB is equal to AD, and AC common, the two fides BA, AC, are equal to the two fides DA, AC, and the bafe BC equal to DC; therefore the angle BAC is equal to DAC; therefore the angle BAD is bifected by the right line a 8. 14 AC. For the fame reafon the angle ADC is bifected by the

right

Book IV. right line DB; but the angles BAD, ADC, DCB, ABC Care equal, therefore their halfs are equal. Again, because BA is equal to AD, and AE common, and the angle BAE to DAÈ, the base BE is equal to ED. For the fame reason, AE is equal to EC: Therefore the right lines AC, BD, are bifected in E; but the two fides AD, DC, are equal to DC, CB"; and the angle ADC to DCB ; therefore the base AC is equal to BD; therefore their halfs are equal: Therefore, if, with the center E, and distance AE, a circle is defcribed, it will pass through the points A, B, C, D,of the fquare: Which was required. Wherefore, &c.

2 II. 2.

b 1:

d 3%. 3. C 32. 3.

T

PRO P. X. PRO B.

O make an ifofceles triangle, having each of the angles at the bafe double to the other angle.

Cut any given right line AB in the point C, fo that the rectangle under AB, BC, be equal to the fquare of AC"; about the center A, and distance AB, describe the circle BDE; in which apply the right line BD equal to AC, and join DA, DC; then the triangle ADB is the ifofceles triangle; having each of the angles at the base BD double the angle at A.

For, defcribe a circle ACD about the triangle ADC.

Then, because the rectangle under AB, BC, is equal to the fquare of AC, and BD is equal to AC, the rec tangle under AB, BC, is equal to the square of BD; therefore BD is a tangent to the circle ACD; therefore the angle CDB is equal to the angle in the alternate fegment CAD. To each add the angle CDĂ; then the whole angle ADB, or its equal ABD f, is equal to the two angles CAD, CDA; but the angle BCD is likewife equal to the angles CDA, CAD; Ax. 1. 1. therefore the angle BCD is equal to CBD; therefore CD is equal to BD; but BD is equal to CA; therefore CD, CA, are equal; therefore the angle CAD is equal to CDA f, but CAD, CDA, are double CAD; therefore ABD or ADB are each double BAD: Which was required.

f 5. 1.

h 6. I.

PRO P. XI. PRO B.

To infcribe an equilateral and equiangular pentagon in a given

circle.

It is required to infcribe an equilateral and equiangular pentagon in the given circle ABCDE.

Book IV,

Make an ifofceles triangle FGH, having each of the angles at the base GH double the other angle F; infcribe the triangle a 10. ADC in the circle ABCDE, equiangular to the triangle FGHb, b 2. then each of the angles ACD, ADC, are double the angle at A; bifect the angles ACD, ADC, by the right lines CE, DB; 9. 3. join AB, BC, DE, EA; then ABCDE is the equilateral and equiangular pentagon required.

For, because the angles ACD, ADC, are bifected by the right lines CE, DB, the five angles DAC, ADB, BDC, DCE, ECA, are equal; therefore the five circumferences AB, BC, CD, DE, AE, are equal d; and the right lines AB, BC, CD, d 26. 3. DE, AE, equal to one another ; the figure is therefore equila- 29. 3. teral; but it is alfo equiangular; for the circumference AB is equal to DE. Add the circumference BCD to both, then the whole circumference ABCD is equal to the whole circumference EDCB; therefore the angle AED is equal to the angle BAE. For the fame reason, the other angles ABC, BCD, f 27. 3. CDE, are equal to BAE, or AED; wherefore the pentagon ABCDE is equilateral, and likewife equiangular. Which was required.

To

PRO P. XII. PRO B.

O defcribe an equilateral and equiangular pentagon about a
given circle.

It is required to defcribe an equilateral and equiangular pentagon about the given circle ABCDE.

Let the points A, B, C, D, E, be the angular points of an equilateral and equiangular pentagon infcribed in the circle; then the circumferences AB, BC, CD, DE, EA, are equal; draw the right lines GH, HK, KL, LM, MG, tangents to the circle in the above points. From F, the center of the 17. 3. circle, draw to the fame points the right lines FB, FC, FD, and join FK, FL; then, because FB, FC, FD, are at right angles to HK, KL, LM, the fquares of FB, BK, are equal to the b 18. 3. fquare of FK, and the fquares of FC, CK, equal to the fquare c 47. I. of FK, and therefore equal to each other d; but the fquares of d Ax. I. I. BF, FC, are equal; which being taken from both, the fquare e def. 15.1. of BK is equal to the square of KC; that is, BK equal to KC. For the fame reason, CL is equal to LD; and because BF, FK, are

equal

f 8. 1.
8 27.3.

Book IV. equal to CF, FK, and the bafe BK equal to KC, the angle BFK is equal to CFK f. For the fame reafon, CFL is equal to LFD; but the whole angle BFC is equal to the whole angle CFD ; therefore the angle KFC is equal to the angle CFL"; and the base KC to CL; but KC is equal to BK; therefore HK is equal to KL. For the fame reafon, KL is equal to LM; therefore the figure is equilateral.

b Ax. 7. 1. i 4. 1.

k Cor. 7.

32. I.

It is likewife equiangular; for the angles B and C are each right ones; and the two angles BFC, BKC, equal to two right angles. For the fame reason, CFD, CLD, are equal to two right angles; but the angles CFD, CFB, are equal; therefore CLD, BKC, are likewife equal. For the fame reason, the whole angle at L is equal to the angle at M; therefore the figure is equiangular; and likewife proved equilateral. Where fore, &c.

210, 1.

b 11. 1.

C 47. 1.

PROP. XIII. PRO B.

To infcribe a circle in a given equilateral and equiangular

pentagon.

It is required to infcribe a circle in the equilateral and equi, angular pentagon ABCDE.

Bifect the fides BC, CD, DE, in the points H, K, L2; from which draw the right lines HF, KF, LF, at right angles to BC, CD, DE ; from the point F, where the right lines HF, KF, interfect each other, draw FC, FD, FE; then, becaufe the angles FHC, FKC, are right angles, the fquare of FC is equal to the fquares of FH, HC, and likewife to the fquares of FK, KC; therefore the fquares of FH, HC, are ed Ax. 1. 1. qual to the fquares of FK, KCd. Take the equal fquare of HC, CK, from both, there remains the fquare of HF equal to the fquare of FK; that is, HF equal FK; therefore HF, FC, are equal to FC, FK, and the bafe HC to CK; therefore the angle HFC is equal to KFC; but the angles FHC, FKC, are right ones; therefore the remaining angle HCF is equal to KCF; therefore the angle BCD is bifected by the right line FC. For the fame reason, FK is equal to FL, and the angle CDE bifected by FD; therefore, becaufe FH, FK, FL, are equal, if, with the center F, and either of thefe diftances, a circle is defcribed, it will touch the fides of the pentagon in the points G, H, K, L, M; wherefore the circle GHKLM is infcribed in the equilateral and equiangular pentagon ABCDE: Which was required.

€ 8. I.