The points M and N are the midpoints of the sides BC and AC of the acute triangle ABC, respectively. There is a point P on AM ... The points M and N are the midpoints of the sides BC and AC of the acute triangle ABC, respectively. There is a point P on AM so that the angles MPC and NPC are equal. Draw a transient line from point B parallel to CP to intersect the NP at point D. Prove that AB = AD.
my try :
at the first Stretching AM from point M to intersect the extend of BD (from D) and I called this point Q after that I tried to use cyclic quadrilateral Properties but I got nowhere
I think this is not hard problem but I stuck on it
 A: $CP$ is the external angle bisector of $\angle APN$ and from here, applying the angle bisector theorem will give $AP=2PN$.
Also, if $PM$ is extended to meet $BD$ at $L$, $PCLB$ is a parallelogram, and from here a bit of angle chasing gives $DP=PL$ and thereafter $DP=2PM$.
$AP=2PN$ and $DP=2PM$ and hence $\triangle APD\sim \triangle NPM$ and therefore $AD=2MN=AB$.
A: 
$CPBP'$ and $CPAP''$ are parallelograms. By angle chase, we can show that the marked angles are all equal and we can show that $ADP'P"$ is an isosceles trapezoid. since $BP'\ \text{parallel}\space AP'' $and $BP'=AP''$, we can conclude that $BAP'P''$ is a parallelogram. So $AD=P'P''=AB$
A: The following develops OP's idea of making use of cyclic quadrilateral/concyclical properties.
Join midpoints $N$, $M$, and extend $CP$, cutting $NM$, $AB$, $AD$ at $G$, $E$, $F$.

By parallels and the angle bisector theorem, in triangles $PNM$, $PAD$,$$\frac{PN}{PM}=\frac{GN}{GM}=\frac{EA}{EB}=\frac{FA}{FD}=\frac{PA}{PD}$$Therefore$$PN\cdot PD=PM\cdot PA$$and $D$ is concyclic with $A$, $N$, $M$.
Next, in triangles $PNG$, $PAF$, since the angles at $P$ are equal, and the angles at $A$, $N$ stand on common arc $MD$, then remaining angles$$\angle NGP=\angle AFP$$Hence since $BD\parallel CF$ and $NM\parallel AB$, then$$\angle ADB=\angle AFP=\angle NGP=\angle ABD$$and$$AD=AB$$
