Dear Mathlinkers, the circumcircle (O) of ABC and the circle "gamma" (1) are tangent at A. Note (2) the circle passing through A, B and P. By a converse of the pivot theorem, (2) is tangent to BC at B and we are done... Sincerely Jean-Louis

First observe that $ADM$ is simmilar to $ABD$ (just angle chasing). Now, let $R$ be the intersection of $BC$ and the reflexion of $AR$ throw $AD$(which is equal to prove that $\angle DAM=\angle RMD$), so the goal now is proving that $MR$ is tangent to $\Gamma$, hence it will be the symmedian of $ADM$. Observe that $\angle PBC=\angle PAB$, let $F=AB\cap \Gamma$, since $FM\parallel BC$, $\angle BMF=\angle MBC$. Since $AR$ is the reflexion of $AP$ across $AD$, then $\angle RAD=\angle DAP$ and since $AD$ is an angle bisector, $\angle FAP=\angle RAM$, therefore the quadrilateral $BAMR$ is cyclic. The angle $\angle PBA=\angle ARM$ and let $x=\angle PAF$, $\angle ARM=\angle ABC-x$, so $\angle AMR=180-(x+\angle ABC-x)=180-\angle ABC$ (that is ovbious since $BAMR$ is cyclic to the oposite angles sum is $180^o$), since $ADM$ is simmilar to $ABD$ the angle $\angle AMD=\angle ADB=180-(\angle ABC+\frac{\angle BAC}{2})$. Hence $\angle DMR=\angle DAM$ so it is a semiinscribed angle, hence $RM$ is tangent to $\Gamma$ and since $BC$ is by construction tangent to $\Gamma$ so $AR$ is the simmedian of $ADM$. So reflecting it on $AD$ will be $AP$ and it is the median.

Let $MN$ cut $AD$ in $L$, $ND$ cut $AP$ in $T$, $BD$ cut $AP$ in $R$ use Pascal's theorem on hexagon $DDAPMN$ then $B, L$ and $T$ are collinear, but it is easy to see that $NL\parallel BC$ because $D$ is midpoint of arc $MN$. Now using the ratios from the paralelism and ceva's theorem on $\triangle ABD$ ceva with point $T$ you get $BR=RD$, showing $AP$ is the median.