The incircle of the acute-angled triangle is tangent to its side at a point . Let be an altitude of triangle , and let be the midpoint of the segment . If is the common point of the circle and the line (distinct from ), then prove that the incircle and the circumcircle of triangle are tangent to each other at the point .

Proof:

Let and let be the antipode of with respect to Let and be the tangency points of with lines and respectively and let . Let be the point at infinity on line

It is clear that so taking perspective at and projecting this harmonic division onto we have that quadrilateral is harmonic. Moreover since the tangents from and to intersect at and since are collinear this implies that quadrilateral is harmonic as well. Since is the tangent from to this means that is on the tangent from to . Therefore are collinear. Therefore .

It is well-known that so the fact that implies that bisects $ \angle{BNC $. Now by Archimedes' Lemma this implies that the circumcircle of is tangent to at as desired, so we are done.

Now I want to discuss the motivation for this solution. We want to prove that two circles are tangent and we have that one of them is tangent to a chord of the other - this immediately means we can rephrase the problem into proving an angle bisection using Archimedes' Lemma. Now, we have a midpoint of an altitude, as well as an angle we want to prove is bisected and these two things imply that we should look at harmonics and projective geometry. When we have an incircle and tangency points and want a harmonic division, constructing is always a good idea. The problem then boils down to proving - in other words, that are colllinear (we phrase it like this because we want to put things in terms of points on ). Now we have lots of tangents and we are thinking about harmonics so finding the right harmonic quadrilaterals is not too arduous a task. All in all, with the right motivation this problem can take 10 minutes! And it's a G7!