Thanks utkarsh for making me a contributor.
here is a problem from Iranian geometry olympiad. It's an easy one and really demonstrates the power of radical axis. In the last part I used coordinates as I was feeling too lazy to use my brain but I think there are several other ways to do it.

Sorry for the edit but couldn't really resist
Nice proof
My proof (second one) demonstrates angle chasing can be used where lots of angles are given



Problem:
Question
is a triangle with and . Let be the midpoint of . Let be a circle passing through tangent in to . Let be the circumcircle of . is intersecting in and in . Prove that is perpendicular to .

Solution 1 (Radical Axis)(Achilles04):
Let be the center of and be that of .
Here we see that ( as and ) so is tangent to . Moreover . So lies on . This shows that as is the diameter of . Also we see that is the radical axis of the two circles. Now if we prove that then we are done. This is easy to prove. For this I used coordinate...
Let be the be the let . So we find that equation of is and that of is , so equation of radical axis we find the slope of this radical axis and see that it is equal to slope of . This proves that or

Solution 2 (Angle Chasing)(utkarshgupta) :
( is tangent to )
But since = Radius of
is equilateral

is perpendicular to
Simple and direct