Let be a triangle with , and let be the foot of the altitude from . Let be a point in the interior of the segment . Let be the point on the segment such that . Similarly, let be the point on the segment such that . Let be the point of intersection of and .

Show that .

Proof:

Let be the circle with center that passes through and let be the circle with center that passes through . Clearly and . Let intersect again at and let intersect again at . Since is on the radical axis of and we have that so by Power of a Point quadrilateral is cyclic. Call its circumcircle and denote its circumcenter by .

Now it is clear that so lies on the polar of with respect to . Moreover since clearly we have that is the polar of with respect to . Therefore is tangent to and similarly is tangent to so we immediately have as desired.

Motivation:

Now, I want to discuss the motivation for this solution, because with this motivation I solved this problem in literally minutes. So first, the length conditions are kind of bad so the constructions of and are natural (in fact, this construction appears often in problems with right triangles). Then since is on the radical axis of these two circles, the construction of is quite motivated as well. If you've drawn a good diagram, the key tangents should then be evident and now given the amount of tangents to circles in the diagram, working with projective geometry should definitely work (and it does!).