93 



(5) The three arcs DJ]C, EBK and DBC cut orthogouallj', two and 

 two, and the common chord of any two of them passes through the center 

 of the third. 



(6) (See Fig. II.) 



Z GPA -i- Z GQA = Z HIF (supplement of Z PGQ). 

 ZGPH = ZGAH. 



ZGQ,F -= ZGAF and adding these three equation-* 

 Z QPI + Z PQI = Z HAF + Z HIF, or 

 180° — Z HIF = Z HAF + Z HIF. 

 180° — 2ZHIF=: ZHAF. But ZH0F = 2ZHIF. 

 .-. 180°— ZHOF= ZHAF and H, A, F, O, are concyclic. 



(7) We have now shown the following points to be concyclic: 



A, G, F, Q— center M. 

 A, G, H, P— center N. 

 A, O, K, Q— center T. 

 A, P, D, O— center S. 

 A, P, I, F— center S'. 

 A, Q, I, H.- center T'. 

 A, H, O, F— center O'. 

 And we will show that X is the center of a circle through A, G, O. I. 



(8) CD, OA and HF are the tliree conmioti chords of circles O, S and O^, 



and must meet in a jjoint. lience HF, the diagonal of FGHI, pnpses 

 through R. 



(9) Since APIF is cyclic ZQAF=ZQIP; and for the same reason ZPAH 



= ZQ[P. .-. ZQ.\F= ZPAH and ZOAF= ZOAH. 

 (10) »Since the circles S', O' and M pass through the points A and F, their 

 centers S', O' and M are in the same line perpendicular to AF. For 

 a similar reason N, ()', T^ are in the same line perpendicular to AH, 

 and S', S,N and T'',T, M are respectively in the same lines perpendicu- 

 lar to PQ or TS. Also T8', TS and MN respectively bisect AI, AO, 

 and AG at right angles. Now the angles SO'S' and SO'N have their 

 sides respectively perpendicular to the sides of the equal angles OAF 

 and O AH. .-. Z SO'S^= Z SO^N and SN = SS^ and in the same way 

 TM^TT^ Hence the lines T^S^ and MN will meet TS at the same 

 point X, and X\ = XG=:XO = XI and X is the center of the circle 

 through A, G, O, I. 

 (11) Now HF, OA, and GI are the three common chords of the 

 circles O, O' and X and must meet in a point. Hence GI the other 



