to the side AB; therefore CBD, or are double of the angle DAC: But DBA is equal to BCD; and conse- BCD is tual to the angles CDA, quently the three angles BDA, DBA, DAC; therefore also BCĎ is double BCD, are equal to one another; and of DAC, and BCD is equal to each because the angle DBC is equal to of the angles BDA, DBA; each, the angle BCD, the side BD is equal therefore, of the angles BDA, DBA (6. 1.) to the side DC; but BD was is double of the angle DAB; whicremade equal to CA; therefore also CA fore av isosceles triangle ABD is deis equal to CD, and the angle CDA scribed, having each of the angles at egual (6. 1.) to the angle DAC; there. the base double of the third angle. fore the angles CDA, DAC together, Which was to be done. PROP. XI, PROB. To inscribe an equilateral and equiangular pentagon in a given cirle. Let ABCDE be the given circle ; sected by the straight lines CE, DB it is required to inscribe an equilateral the five angles DAC, ACE, ECD and equiangular pentagon in the circle CDB, BDA are equal to one another, ABCDE. but equal angles stand upon equal Describe (10. 4.) an isosceles tri- (26. 3.) circumferences; therefore the angle FGH, having each of the angles five circumferences AB, BC, CD, DE, at G, H, double of the angle at F; EA are equal to one another; and and in the circle ABCDE inscribe cqual circumferences are subtended (2. 4.) the triangle ACD equiangular by equal (29. 3.) straight lines ; thereto the triangle FGH, so that the angle fore the five straight lines AB, BC, CAD be equal to the angle at F, and CD, DE, EA are equal to one ano ther. Wherefore the pentagon ABCDE is equilateral. It is also equiaugılar; because the circumference AB is equal to the circumference DE: If to each be added BCD, ihe whole ABCD is equal to the whole EDCB; And the angle AED stands on the circumference ABCD, and the angle BAE on the circumference EDCB ; thereeach of the angles ACD, CDA equal fure the angle SAE is equal (27. 3.) to the angle at G or H; wherefore to the angle AED: For the same reaeach of the angles ACD, CDA is son, each of the angles ABC, BCD, double of the angle CAD. Bisect CDE is equal to the angle BAE, or (9. 1.) the angles A CD, CDA by the AED: Therefore the pentagon ABCDE straight lines ČE, DB: and joiò AB, is equiangular; and it has been shewn BC, DE, EA. ABCDE is the pen- that it is equilateral. Wherefore, in tagon required. the given circle, an equilateral and Because each of the angles ACD, equiangular pentagon has been inCDA is double of CAD, and are bi- scribed. Which was to be done, : PROP. XII. PROB. To describe an equilateral and equiangular pentagon about a given circle. Let ABCDE he the given circle ; ral and equiangular pentagon about it is required to describe an equilate. the circle ABCDE. Let the angles of a pentagon, in- ence BC is equal to the circumferscribed in the circle, by the last pro- ence CD, the angle BFC is equal position, be in the points A, B, C, D, (27. 3.) to the angle CFD: and BFC E, so that the circumferences AB, is double of the angle KFC, and BC, CD, DE, EA, are equal; (11. 4.) CFD double of CFL; therefore the and through the points A, B, C, D, E, angle KFC is equal to the angle draw GH, HK, KL, LM, MG, touch- CFL; and the right angle FČK ing (17. 3.) the circle ; take the centre is equal to the right angle FCL: F, and join FB, FK, FC, FLCD: Therefore, in the two triangles FKC, And because the straight line KL FLC, there are two angles of one touches the circle ABCDE in the equal to two angles of the other, each point C, to which FC is drawu from to each, and the side FC, which is the centre F, FC is perpendicular adjacent to the equal angles in each (18. 3.) to KL; therefore each of is common to both; therefore the the angles at C is a right angle: other sides shall be equal(26. 1.) to the For the same reason, the angles other sides, and the third angle to the at the points B, D are right angles; third angle: Therefore the straight and because FCK is a right angle, line KC is equal to CL, and the angle the square of FK is equal (47. 1.) FKC to the angle FLC : And because to the squares of FC, CK: For KC is equal to CL, KL is double of the same reason, the square of FK KC: In the same manner, it may be is equal to the squares of FB, BK: shewn that HK is double of BK: And Therefore the squares of FC, CK are because BK is equal to KC, as was equal to the squares of FB, BK, of demonstrated, and that KL is double which the square of FC is equal to of KC, and HK double of BK, HK the square of FB; the remaining shall be equal to KL: In like manner square of CK is therefore equal to the it may be shewn that GH, GM, ML remaining' square of BK, and the are each of them equal to HK or KL: straight line CK equal to BK: And Therefore the pentagon GHKLM is because FB is equal to FC, and FK equilateral. It is also equiangular; common to the triangles BFK, CFK, for, since the angle FKC is equal to the two BF, FK are equal to the two the angle FLC, and that the angle CF, FK; and the base BK is equal HKL is double of the angle FKC, to the base KC; therefore the angle and KLM double of FLC, as was BFK is equal (8. 1.) to the angle before demonstrated, the angle HKL KFC, and the angle BKF to FKČ; is equal to KLM: And in like manwherefore the ner it may be shewn that each of angle BFC is the angles KHG, HGM, GML is double of the equal to the angle HKL or KLM: angle KFC, & Therefore the five angles GHK, HKL, BKC double of KLM, LMG, MGH being equal to FKC: For the one another, the pentagon GHKLM same reason, is equiangular : And it is equilateral, the angle CFD as was demonstrated ; and it is des is double of scribed about the circle ABCDE, the angle CFL, and CLD double of which was to be done. CLF: And because the circumfer G E H M B D K PROP. XIII. PROB. To inscribe a circle in a given equilateral and equiangular pentagon. Let ABCDE be the given equila- ired to inscribe a circle in the teral and equiangular pentagon ; it is pentagon ABCDE. ELEMENTI OF RUCLID. a G B B Bisect (9. 1.) the angles BCD, EA : And because the angle HCF is cCDE by the straight lines CF, DF, qual to KCF,and the right angle FHC and from the point F, in which they equal to the right angle FKC; in meet, draw the straight lines FB, FA, the triangles FHC,FKC, there are two FE: Therefore, since BC is equal tó angles of one equal to two angles of CD, and CF common to the triangles the other, and the side FC, which is BCF, DCF, the two sides BC, CF opposite to one of the equal angles in are equal to the two DC, CF; and each, is common to both; therefore the angle BCF is equal to the angle the rather sides shall be equal, (26. 1.) DCF: therefore the base BF is equal each to each; wherefore the perpen(4. 1.) to the base FD, and the other dicular FH is equal to the perpendiangles to the other angles, to which cular FK: In the same manner it the equal sides are opposite ; there may be demonstrated, that FL, FM, fore the angle CBF is equal to the FG are each of them equal to FH or angle CDF: And because the angle FK: Therefore the five straight lines CDE is double of CDF, and that CDE FG, FH, FK , FL, FM are equal to is equal to CBA, and CDF. to CBF; one another: Wherefore the circle CBA is also double of the angle CBF; described from the centre F, at the therefore the angle ABF is equal to distance of one of these five, shall the angle CBF; wherefore the angle pass through the extremities of the ABC is bisected by the straight line other four, and touch the straight BF: In the same manner it may be lines AB, BC, CD, DE, EA, because demonstrated, that the angles BAE, the angles at the points G, H, K, L, AED, are bic M are right angles ; and that a sected by the straight line drawn from the extremistraight lines ty of the diameter of a circle at right AF,FE: From angles to it, touches (16. 21.) the the point F circle : Therefore each of the straight draw (12. 1. lines AB, BC, CI), DE, EA touches FG, FH, FK, the circle ; wherefore it is inscribed FL, FM per in the pentagon ABCDE. Which pendiculars to was to be done. the straight lines AB, BC, CD, DE, PROP. XIV. PROB. To describe a circle about a given equilateral and equiangular pentagon. Let ABCDE be the given equilate. position, that the angles CBA, BAE, ral and equiangular pentagon ; it is AED 'are_bisected by the straight required to describe a circle about it. lines FB, FA, FE: And because the Bisect (9. 1.) the angles BCD, angle BCD is equal to the angle CDE by the straight lines CF, FD, CDE, and that FCD is the half of and from the point F, in which they the angle BCD, and CDF the half meet, draw the of CDE; the angle FCD is equal to straight linesFB, FDC; wherefore the side CF is equal FA, FE, to the (6. 1.) to the side FD: In like manpoints B, A, E. ner it may be demonstrated that FB, It may be de FA, FE, are each of them equal to monstrated, in FC or FD: T'herefore the five straight lines FA, FB, FC, FD, FE, ner as in the are equal to one another; and the preceding pro circle described from the centre F, H с K B E the same man at the distance of one of them, shall the equilateral and equiangular penpass through the extremities of the tagon ABCDE. Which was to be other four, and be described about done. PROP. XV. PROB. To inscribe an equilateral and cquiangular hexagon in a given circle. Let ABCDEF be the given circle : to one another: But equal angles it is required to inscribe an equilate- stand upon equal (26. 4.) circumfer. ral and equiangular hexagon in it. ences; therefore the six circumferen Find the centre G of the circle ces AB, BC, CD, DE, EF, FA are ABCDEF, and draw the diameter equal to one another : And equal cir. AGD; and from D as a centre, at cumferences are subtended by equa the distance DG, describe the circle (29. 3.) straight lines ; therefore the EGCH, join EG, CG, and produce six straight lines are equal to one anthem to the points B, F; and join other, and the hexagon ABCDEF is AB, BC, CD, DE, EF, FA: The equilateral. It is alɛo equiangular; hexagon ABCDEF is equilateral and for since the circumference AF is equiangular. equal to ED, to each of these add Because G is the centre of the circle the circumference ABCD; therefore ABCDEF, GE is equal to GD: And the whole circumference FABCD because D is the centre of the circle shall be equal to the whole EDCBA: EGCH, DE is equal to DG; where. And the angle FED stands upon th. fore GE is equal to ED, and the tri- circumference FABCD, and the angla angle EGD is equilateral ; and there- AFE upon EDCBA; therefore the fore its three angles EGD, GDE, angle AFE is equal to FED: In the DEG are equal to one another, be- same manner it may be demonstratec cause the angles at the base of an that the other angles of the hexagor isosceles triangle are equal; (5. 1.) ABCDEF are each of them equal to and the three angles of a triangle are the angle AFE or FED: Therefore equal (32. 1.) to two right angles; the hexagon is equiangular; and it therefore the angle EGD is the third is equilateral, as was shewn; and it is part of two right angles : In the same inscribed in the given circle ABCDEF manner it may be de Which was to be done. monstrated, that the Cor. From this it is manifest, that angle DGC is also the the side of the hexagon is equal to the third part of two right straight line from the centre, that is, angles: And because to the semi-diameter of the circle. the straight line GC And if through the points A, B, C, makes with EB the D, E, F, there be drawn straight adjacent angles EGC, lines touching the circle, an equilate CĞB equal (13. 1.) to two right ral and equiangular hexagon shall be angles ; the remaining angle CGB is described about it, which may be dethe third part of two right angles; monstrated from what has been said therefore the angles EĞD, VGC, of the pentagon ; and likewise a circle CGB are equal to one another : And may be inscribed in a given equilato these are equal (15. 1.) the verti- teral and equiangular hexagon, and cal opposite angles BGA, AGF,FGE: circumscribed about it, by a method Therefore the six angles EGD, DGC, like to that used for the pentagon. CGB, BGA, AGF, FGE, are equal PROP. XIV. PROB. To inscribe an equilateral and equiangular quindecagon in a given circle. Let ABCD be the given circle; it same parts : Bisect (30. 3.) BC in E; is required to inscribe an equilateral therefore BE EC are, each of them, and equiangular quindecagon in the the fifteenth part of the whole circumcircle ÅBCĎ ference ABCD: Therefore, if the Let AC be the side of an equilateral straight lines BE, EC be drawn, and triangle inscribed (2. 4.) in the circle, straight lines equal to them be placed and AB the side of an equilateral and (1. 4.) around in the whole circle, an equiangular pentagon inscribed (11.4.) equilateral and equiangular quindecain the same, therefore, of such equal gon shall be inscribed in it. Which parts as the whole circumference was to be done. ABCDF contains fifteen, the circum- And, in the same manner as was ference ABC, being the third part of done in the pentagon, if, through the the whole, con points of division made by inscribing tains tive; and the quindecagon, straight lines be the circumfer drawn touching the circle, an equilaence AB, which teral and equiangular quindecagon is the fifth part shall be described about it: And likeof the whole, wise, as in tie pentagon, a circle may contains three; be inscribed in a given equilateral and therefore BC, equiangular quindecagon, and cirtheir difference contains two of the cumscribed about it. E |