64 Strong’s Problems, 
FEG. Wherefore their halves FNG, HKI are equal. 
Consequently NG, IK H are similar, and likewise the seg- 
ments FAG,IBH. Therefore a poimt C is found as re- 
quired.— @. E. I. 
Cor. {.. By a similar construction, similar segments aa 
be cut from spheres given in position and magniiude by a 
plane, as is manifest from the solution of this Problem. 
Cor. If. When in Case J. the circles approach to equality, 
the point A becomes infinitely distant, and the line AC be- 
comes parallel to Al, which passes through the centres of 
the circles. 
Cor.1tf. When in Case II. the circles hence equal, 
the peint C (as in Case II. Prob. VIL.) is equidistant from 
the centres of the circles. 
Cor. 1V. In Case I. the points C, I, M, O, are in the 
circumference of a circle. For 4FCI = ZGNL and Z4GNR 
= £§CH, therefore the whole angle ICH = whole angle 
-LNO. But ZLNO+ZOML—two right angles, therefore 
ZICH +ZOML = two right angles. Therefore the remain- 
ing an; Bes CIM, COM = two right angles. ‘Therefore the 
poiats ©, F, M, 0, are in the circumference of a circle.— 
In hike manner H, K, L, N are in the circumference of a 
circle. "Therefore the rectangle AM. AI = AC. AO, and 
also AK. ALL = AH. AN. 
Cor. V. Because (in Case I.) the segment yFAG is 
similar to the segment I[BHza, the angle [Hx = GF y and 
the angles at C being vertical are equal; therefore the tri 
angles CyF, CHa are similar. But the triangle CrI is sim- 
ilar to the triangle CHz. For the angles Irv+1Hax =two 
right angles; and Ivw+Irc = two right angles: taking from 
both, the common angle Irv, there remains Cr] — CHa, 
and the angle at'C being common to the two triangles they 
are similar. Hence CFy and CH being similar and like- 
wise CHa and Crl, CF y is similar to Crl. Therefore C1: 
Cn: Cy: CE. Therefore Cl, CF = Cr. Cy. There- 
fore the points I, r, F, y are in the circumference of a circle. 
fn like manner it may be shewn that the points G, 2, H, z 
are in the circumference of a circle. 
(To be continued.) 
