574 



MATHEMATICS. PLANE GEOMETRY. [BOOK iv. PROP. TIL XIL 



the figure it rtctanyular, .'. it ii a $quan ; and it it in- 

 teribedin thteirtU ABCD. Which wai to be done. 



PROPOSITMN VIL PROBLEM. 

 To dtterUt fftMr* <fcm/ a girt* eirelt (ABCD). 

 Draw two diameters, AC, B D, at right angles to one 

 another, and through A, B, C, D, draw FG, G H, H K, 

 K F, touching the circle : * the 



:u. figure O H K F shall 

 be the square required. Because 

 F G touches the circle, and E A 

 is drawn from the centre to the 

 point of contact A, the angles at 



t is in. A are right angles, t 

 For a like reason, the angles at 

 B, C, 1), are right angles. And 

 IH . .itiM'th- .in.. A V. I'.i-a rs.'iit 

 angle, u likewise E B G, G H is 



r* I. parallel to A C ; * for a like reason A C is 

 parallel to F K. In a similar manner may it be proved 

 that G F, H K. are each parallel to B 1 ), . . the figures 

 G K, GO, A K, FB, B K, are parallelograms, . . G F 

 + w I. - H K, and G H = F K ; t and because A C 



- B 1), and that A G = G H = F K : and B D = (i F 



- HK, /. GH, FK, are each = G F or H K, .-. the 

 figure F G H K is trilateral. It is also rectangular ; 

 for G E being a parallelogram, and A E B a right angle, 



jii. G is a right angle ; * and G K, being a 

 parallelogram with a right angle at G, ia rectangular,^ 



+ Cor. u i. .'. it is r square ; and it i* detcribed about 

 the circle ABCD. Which was to be done. 



PROPOSITION VIIL PROBLEM. 



To inter it f a circle in a given tquare (ABCD). 



10 I. Bisect each of the sides A B, A D in F, E ;* 

 draw E H parallel to A B or DC, and F K parallel to 



tail. A D or B C ; t then, if with their point of 

 intersection G, as centre, and G F, or G E, as a radius, 

 a circle be described, it will be that required. 



Each of the figures A K, K B, A H, H D, A G, G C, 

 A K t) B G, G D, being a parallelo- 



gram, their opposite sides are 

 S4 1. equal :* and because 

 A D = A B, and A E=half A D, 

 and A F=lialf AH, A E = A F, 

 . . the sides opposite to these 

 are equal viz. , F G = G E. In 

 like manner, it may be demon- 

 strated that GII = GK, and 

 GH = GF, .-. the four GE, 

 OF, OH, G K are equal to 

 one another ; and the circle described from centre G, 

 with any one of them for a radius, will pass through tin: 

 extremities of all, and will touch A B, B C, C \), D A, 

 because the angles E, F, H, K are right angles ;t 

 + la in. cor. the circle it .'. inscribed in the square 

 ABCD. Which was to be done. 



PROPOSITION IX. PROBLEM. 

 To detcribe a circle about a given iquare (ABC D). 

 Draw A C, B I), intersecting in E : with E as centre, 

 and K A as radius, describe a circle : it will bo that 

 required. 



Because D A - A B, and A C is common to the tri- 

 angles D A C, B A I', the two sides 

 D A, A C-the two 1! A, A C, each 

 to each; and DC-BC, .-. the 

 8 1. angle DAC-BAC;* 

 that is, the angle D A B is bisected 

 by A C. In like manner, it may 

 be proved that the angles A H C, 

 BCD, CD A, are severally bi- 

 sected by B 1), C A. Again : the 

 angle D A B- A B C, also E A B- 

 half D A B, and E B A - half A B C, .-. E A B - E B A, 



1 1. /. E A - E B.t In like manner, it may be de- 

 moniitratcd that E C, ED-EB, E A, each to each, .-. 

 the four E A, E B, E C, E D are equal to one another ; 



C 



and the circle whose centre is E, and radius E A, passes 

 tlirough the extremities of all, and it .: detcribed about 

 the tquart ABCD. Which was to be done. 



PROPOSITION X. PROBLEM. 



To detcribe an itotcelet triangle having each of the anglet 

 at the bate double of the third angle. 



Take any straight line AB, and divide it in C, so 

 . 11 ii. that A B B C - A C 1 ,* and with centre A and 

 radius A B, describe the circle B D E, in which place 

 + i IV. B D= A C ;t and draw A D ; the triangle A B D 

 shall be such that each of thu 

 angles A B D, A D B shall be 

 double of B A D. 



Draw D C ; and about the 

 triangle A C D describe the 

 circle A C D ; then because 



Conn. A B-B C A C 1 ,* 

 and AC=BD, .-. AB-BC- 

 BD 3 , .-. BD touchettbe cir- 



is;in. clef AC Din D, .-. 

 the angle BDC-DAC;* 



si ill. to each of these add 

 C D A, . '. B D A = D A C + C D A ; but the exterior 



+ 3ZI. angle BCD = DAC +CDA,t .'. BDA- 



31. BCD; but BDA=B, because AB = AD,.-. 

 B=B C D, . . the three angles DBA, B D A, BCD, 

 are equal to one another. Again, because the angle 



.81 DBC = DCB, DC = DB;* but DB was 



made - C A, /. C A = C D, .-. the angle C D A - 



tsi C AD,t .-. CD A -f C AD = twice CAD ; 



sal. butBCD=Cl) A + CAD,* .-. BCD=twice 

 CAD; but B C D = B D A = B, as before proved, . . 

 each of the angles B D A, DBA is double of BAD, 



on isosceles triangle A B D has f>een described, having 

 each of the anglet at the bate double of the. third angle, 

 Which was to be done. 



PROPOSITION XI. PROBLEM. 



T i inscribe an eguiltiteral and equiangular pentagon in a 

 yiten circle (ABC D). 



Describe an isosceles triangle F G H, having each of 

 10 iv. the angles G, H, double of F ;* and in the 

 circle inscribe a triangle A C D, equiangular to the 



triangle F^G H, so tlmt 

 the angle CAD may be 

 + 2 iv. - F ;t then 

 each of the angles A CD, 

 ADC is double of CAD. 

 Bisect A CD, ADC, by 

 the straight lines C E, 

 91. DB ;* and draw 

 AB, BC, DE, EA. 

 ABCDE shall be the 

 pentagon required. Because each of the angles A C D, 

 A D C is double of C A D, and that they are bisected by 

 C E, D B, . ' . the five angles DAG, ACE, BCD, CDB, 

 BD A are equal to one another ; .-. the five arcs AB, 

 + 28 i ii. B C, C D, D E, E A are equal to one another, t 

 . the lines A B, B C, C D, D E, EA are equal to one 



win. another,* .-.the pentagon ABODE is equila- 

 teral. It is also equiangular, for since the arc A B 

 the arc D E, if to each B C D be added, the whole 

 ABCD=EDCB; but these arcs subtend the angles 



AED, BAE, .-. AED-BAE.t Fora 

 like reason each of the angles A B 1C, B C D, C D E - 

 B A E or A E D ; . . the pentagon is equiangular, . . in 

 the given circle an equilateral and equiangular pentagon 

 hat been detcribed. Which was to be done. 



PROPOSITION XII. PROBLEM. 



To detcribe an equilateral tHdtqui<ni<inl<ir pentagon aliout 

 a given circle (ABCD). 



Let the vertices of a pentagon, inscribed in the circle, 

 by last proposition, be at A, B, C, D, E ; so that the 

 arcs A B, BC, CD, DE, E A are equal ; and through 

 these points draw G H, H K, K L, L M, MG, touching 



