An elementary treatise on modern pure geometry (1893)
Lachlan, Robert · London, New York, Macmillan and Co. · leaf 152
if we divide thelines BB, CC in the points K, L, so that BK : KB = CL : LC = BG : BC,it is evident that KL must be the axis of similitude. Again the triangles BOC, BOC are inversely similar, so thatthe perpendiculars from 0 on BC, BC must be in the same ratioas BC : BC. Consequently if BC, BC… … It is obvious that when 0 has been found in this way, thetriangles OAB, OAC are inversely similar to the triangles OAB,OAG; and that KL bisects the angle AOA. Hence 0 is thecentre, and KL the axis of similitude of the triangles. 212. Ex. ]. Find the centre and axis of similitude of the triangles…
Geometry
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