Q:

Geometry problem (Iran Olympiad)

Let $\triangle ABC$ be any triangle. Suppose the angle bisector of $\angle BAC$ intersects $BC$ at $D$. Let $\Gamma$ be a circle tangent to $BC$ at $D$ and so that $A$ belongs to the circumference of $\Gamma$. If $M$ is the (second) intersection point of $AC$ and $\Gamma$, and if $BM$ intersects $\Gamma$ at $P$, then prove that $AP$ must be a median of $\triangle ABD$.

I need some help with this problem. It was taken from an Iran Math Olympiad (from 1999 I believe).
I've mainly tried proving that $BD \over JD$$=2$ (here $J$ is $AP \cap BC$), by various methods. First, using power of points and the angle bisector theorem, no success. Then I've tried some angle chasing to find some similar triangles, just to find the same relations I had found using power of points...
Finally I went full trigonometry over it, but then again I was never quite good at trigonometry anyway, so I couldn't get very far.
I would love some hints, because I'm pretty much stuck. Thanks in advance.

A:

P is not a nice point, so let's try and avoid it. 
Hint: show that $JB^2=JP \times JA = JD^2$.
Hint: Show that 2 triangles are similar to prove the first. 
Hint: define N in a similar manner to M

