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$$=2$ (here $J$ is $AP \cap BC$), by various methods. First, using power of points and the angle bisector theorem, no success. Then I've tried some angle chasing to find some similar triangles, just to find the same relations I had found using power of points...
Finally I went full trigonometry over it, but then again I was never quite good at trigonometry anyway, so I couldn't get very far.
I would love some hints, because I'm pretty much stuck. Thanks in advance.
 A: P is not a nice point, so let's try and avoid it. 
Hint: show that $JB^2=JP \times JA = JD^2$.
Hint: Show that 2 triangles are similar to prove the first. 
Hint: define N in a similar manner to M
A: For completeness, I'll post my solution here.
By looking at the power of $J$ with respect to $\Gamma$, we see that:
$|JD|^2 = |JP|\cdot|JA|$.
On the other hand, let $\angle BAD = \alpha$ and $\angle PAD = \beta$. From this we can see that $\angle BAJ = \alpha - \beta$. It is also true that $\angle DAM = \alpha$ (since $AD$ is the angle bisector of $\angle BAM$). So we have $\angle MBD = \angle DAM - \angle PAD = \alpha - \beta$. It follows that $\triangle BJP \sim \triangle AJB$, whence: $$\frac{|JB|}{|JA|}=\frac{|JP|}{|JB|} \Rightarrow |JB|^2 = |JA| \cdot |JP|$$
So $|JD|^2 = |JP| \cdot |JA| = |JB|^2 \Rightarrow |JD| = |JB|$, which is what we wanted to prove. $\blacksquare$

A: See the Deathkamp Drone's picture.
$$\measuredangle MBC=\measuredangle AMB-\measuredangle C=\measuredangle AMD-\measuredangle PMD-\measuredangle C=\measuredangle ADB-\measuredangle PAD-\measuredangle C=$$
$$=\measuredangle DAC-\measuredangle PAD=\measuredangle BAD-\measuredangle PAD=\measuredangle BAP,$$
which says $$\Delta BPJ\sim\Delta ABJ,$$ which gives
$$BJ^2=JP\cdot JA=JD^2$$ and we are done!
