64 | Strong’s Problems. 
FEG. Wherefore their halves FNG, HKI are equal. 
a res FING, IKH are similar, and likewise the seg- 
s FAG, IBH. Therefore a point C is found as re- 
fa 
or. I. By a similar construction, similar segments may 
be cut from spheres given in position and magnitude by a 
plane, as is manifest from the solution “of this Problem. 
Cor. I. When in Case I. the circles approach to equality, 
the point A becomes infinitely distant, and the line AC be- 
comes parallel to AJ, which passes through the centres of 
the circles. 
Cor. THI. When in Case II. the circles become equal, 
the point C (as in Case II. Prob. VIL.) is equidistant from 
the centres of the circles. 
Cor. IV. In Case I. the points C,I, M, O, are in the 
circumference of a circle. For FCI = ZGNL and ZGNR 
ae TAG a: ooo the whole eet ICH = whole angle 
ES 
ing angles CIM. COM — eee ret. Bigles. Therefore the 
; pega tore tenia s 
In like manner , L, N are in the circumference of a 
circle. Therefore the rectangle AM. AI = A , and 
Iso AK. H. AN, 
— 
=— 
Cor. V. Because (in Case II.) the segment yFAG is 
similar to the segment IBHz, the angle {He = Fy and 
the angles at C being vertical are equal; therefore the tri- 
angles CF, ‘CH are similar. But the triangle CrI is sim- 
ilar to the triangle CHz. For the angles Irr+ITHz =two 
tight angles ; and Ire +Ire = two right angles: taking from 
“both, the common angle Ira, there remains Cri = A 
and the angle at C being common to the two triangles they 
are similar. Hence Cy and CHv being similar and like- 
wise CHa and CrI, wad Ae is similar to Cr. nie A CI: 
r::Cy:CF. Therefore Cl. CF = Cr. Cy. There- 
fore the points I, r, F, y are in the circumference of a circle. 
In like manner it may be shewn that the points G, z, H, + 
are in the circumference of a circle. 
(To be continued.) 
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