#Similarity
3 messages · Page 1 of 1 (latest)
I think I have a solution, actually. Just draw an arbitrary circle that is tangent to the lines, which let's say intersect at O. If we draw the line PO, that line will intersect our arbitrary circle at two points.
Say P'1 and P'2. We can construct two similar circles then by scaling our arbitrary circle that go through P. The ratios will be OP/OP'1 and OP/OP'2.