Book IV. It is required to describe a circle about the triangle ABC. Bifect the sides 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 sides BC, or without the triangle, in the point F. In either case, because AD is equal to DB, and DF common, the two fides AD, DF, are equal to BD, DF; and the angle BDF to ADFb; therefore the base BF is equal to the base AFC, For the same reason AF is equal to FC: Therefore, if, with the center F, and either of the distances AF, FB, or FC, a circle is described, it will touch all the angles of the triangled: Which d Def. 6. was required. It is required to inscribe a square in the given circle ABC. Draw the diameters BD and AC at right angles to, and bi. fecting each other in E; join BA, AD, DC, CB; then ADCB is a square. For the two sides BE, EA, are equal to DE, EA?, and the angle BEA to AED”; therefore the base BA is equal to AD. For the same reason AD is equal to DC, and DC to CB ; therefore the four fides are equal : But the angle BAD is a right angles; therefore ADC, DCB, ABC, are each right angles"; therefore the figure is a squared: Which was requid def. 30. 1. red. a II. 1. b 4. I. € 31. 3. 2 II. I. It is required to describe a square 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 GAĚ, AEB, are right angleso, GF is parallel to BDC. For the same reason HK is parallel to BD; therefore GF is parallel to HKC, For the same reason, GH is b 16. 3. C29. I, d 30. I. parallel to FK; therefore GHKF is a parallelogram; but GF, Book IV, BD, are equal, and likewise BD equal to HK, therefore GF is equal to HK f. For the same reason, GH is equal to FK; fore 34. 1. each is equal to AC; but AC is equal to BD; therefore GH is f Ax. 1. 1. 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; b 16.30 therefore all the 'angles are right ones; therefore GHKF is a square, described about the circle ABCD: Which was to be done. It is required to inscribe a circle in the square ABCD. Bifect the fides AB, AD, in the points E, Fa; through Ea 10.10 draw EH parallel to AB, and through F draw FK parallel to b 31. 1. AD or BC b; then AE, FG, are equal to one another', and C 34. 5. likewise ED, GK; but AE is equal to ED"; therefore FG is equal to GK; but AF is equal to AE d; therefore EG is equal d Ax. 7. I. to FG, and FB to GH"; therefore GE, GF, GH, GK, are equal; therefore if, with the center G, and either of these distances, a circle is described, it will touch the square in the points E, F, H, K; for if not, let it cut the square, then a right line drawn at right angles to the diameter of a circle will fail within the circle; which cannot be ®; therefore the circlee 16. 3, EFHK is inscribed in the square ABCD: Which was required. PRO P. IX. PROB. To describe a circle about a given fquare. It is required to describe a circle about the given square ABCD. Join AC, BD, mutually cutting each other in the point E. For, because AB is equal to AD, and AC common, the two ķides BA, AC, are equal, to the two sides DA, AC, and the base 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 same reason the angle ADC is bisected by the right ود اظ Book IV. right line DB ; but the angles BAD, ADC, DCB, ABC are equal, therefore their halfs are equal. Again, because BA . is equal to AD, and AE common, and the angle BAE to DAE, the base BE is equal to ED. For the same reason, AE is equal to EC: Therefore the right lines AC, BD, are bisected in E, but the two sides AD, DC, are equal to DC, CB "; and the angle ADC to DCB b; therefore the base AC is equal to BDC; therefore their halfs are equal : Therefore, if, with the center E, and distance AE, a circle is described, it will pass through the points A, B, C, D,of the square : Which was re. quired. Wherefore, &c. PRO P. X. PRO B. T! O make an isosceles triangle, having each of the angles at the base double to the other angle. 1l. 2 bi: Cut any given right line AB in the point C, so that the rectangle under AB, BC, be equal to the square of AC"; about the center A, and distance AB, describe the circle BDE ; in which apply the right line BD equal to AC 6, and join DA, DC; then the triangle ADB is the isosceles triangle; having each of the angles at the base BD double the angle at A. C 5. For, describe a circle ACD about the triangle ADC°. Then, because the rectangle under AB, BC, is equal to the square of AC, and BD is equal to AC, the rectangle 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 segment CAD. To each add the angle CDA ; then the whole angle ADB, or its f s. 1. equal ABD f, is equal to the two angles CAD, CDA; but the angle BCD is likewife equal to the angles CDA, CAD"; Ax. I. 1. therefore the angle BCD is equal to CBD 8; therefore CD is equal to BDh; but BD is equal to CA; therefore CD, CA, are equal 8 ;, 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. d 37, 3: C 32. 3. h 6.1. PRO P. XI. PRO B. b It is required to inscribe an equilateral and equiangular pen. Book IV, tagon in the given circle ABCDE. Make an isosceles triangle FGH, having each of the angles at the base GH double the other angle F'; inscribe the triangle a 10. ADC in the circle ABCDE, equiangular to the triangle FGHb; then each of the angles ACD, ADC, are double the angle at A; bisect 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 bisected 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 also 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 BAEf. For the same reason, the other angles ABC, BCD, f 27. 3. CDE, are equal to BAE, or AED; wherefore the pentagon ABCDE is equilateral, and likewise equiangular. Which was required. PRO P. XII. PRO B. To describe an equilateral and equiangular pentagon about a given circle. It is required to describe 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 inscribed 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 thea 17. 3. circie, draw to the same 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 squares of FB, BK, are equal to the b 18. 3. square of FK, and the squares of FC, CK, equal to the square c 47. 1. of FK, and therefore equal to each other d; but the squares of a Ax. 1.*• BF, FCR, are equal; which being taken from both, the square - def. 15.1! of BK is equal to the square of KC ; that is, BK equal to KC. For the same reason, CL is equal to LD; and because BF, FK, are equal f 8. 1. Book IV. equal to CF, FK, and the base BK equal to KC, the angle BFK is equal to CFK f. For the same reason, CFL is equal to CFD 8; therefore the angle KFC is equal to the angle CFL"; i 4. 1. and the base KC to CL i; but KC is equal to BK; therefore HK is equal to KL. For the same reason, KL is equal to LM; therefore the figure is equilateral. It is likewife equiangular; for the angles B and C are each right ones; and the two angles BFC, BKC, equal to two right k Cor.7. anglesk. For the same reason, CFD, CLD, are equal to two right angles; but the angles CFD, CFB, are equal; therefore CLD, BKC, are likewise equal. For the same reason, the whole angle at L is equal to the angle at M; therefore the figure is equiangular; and likewise proved equilateral. Where, fore, &c, 32, 1. PRO P. XIII, PRO B. To infcribe a circle in a given equilateral and equiangular pentagon. C47. 1. It is required to inscribe a circle in the equilateral and equi. angular pentagon ABCDE. Bifect the sides BC, CD, DE, in the points H, K, L'; from which draw the right lines HF, KF, LF, at right angles bol. 1. to BC, CD, DE b; from the point F, where the right lines HF, KF, intersect each other, draw FC, FD, FE ; then, because the angles FHC, FKC, are right angles, the square of FC is equal to the squares of FH, HCC, and likewise to the squares of FK, KC; therefore the squares of FH, HC, are e. d Ax. 1. 1. qual to the squares of FK, KCd. Take the equal square of HC, CK, from both, there remains the square of HF equal to the square of FK; that is, HF egual FK; therefore HF, FC, are e qual to FC, FK, and the base HC to CK; therefore the angle ¢ 8. 3 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 bisected by the right line FC. For the same reason, FK is equal to FL, and the angle CDE bifected by FD; therefore, because FH, FK, FL, are equal, if, with the center F, and either of these distances, a circle is described, it will touch the sides of the pentagon in the points G, H, K, L, M; wherefore the circle GHKLM is inscribed in the equilateral and equiangular pentagon ABCPE: Which was ren quired. |