geometry might be easy

[SANZEED]

Extend side $CB$ of triangle $ABC$ beyond $B$ to a point $D$ such that $DB=AB$. Let $M$ be the midpoint of side $AC$. Let the bisector of $∠ABC$ intersect line $DM$ at $P$. Prove that $∠BAP =∠ACB$.

[photon]

let $BP$ intersects $CA$ at $X$.by angle bisecting theorem,
$\frac{CB}{AB}=\frac{CX}{AX}\Rightarrow \frac{CB}{DB}=\frac{CX}{AX}$
so ,$BP,AD$ are parallel. $\angle BPD=\angle ADP,\angle ADC=\angle PBC=\angle ABP$
In $\Delta ADC$ , $\Delta ADM,\Delta MDC$ have same area .
$ \frac{1}{2}CD.DM.sin\angle CDM=\frac{1}{2}AD.DMsin\angle ADM$
$\Rightarrow \frac{sin\angle CDP}{sin\angle ADP}=\frac{AD}{CD}$
$\Rightarrow \frac{sin\angle BDP}{sin\angle BPD}=\frac{AD}{CD}$
$\Rightarrow \frac{BP}{BD}=\frac{AD}{CD}$
$\Rightarrow \frac{BP}{AB}=\frac{AD}{CD}$
and $\angle ADC=\angle ABP$,$\Delta BAP\sim \Delta ACD$
So, $\angle BAP=\angle ACD$

[SANZEED]

My solution is euclidean indeed:
Construct line \[BF||CA\] with $F$ on line $AD$. Let $DM$ intersect $BF$ at $E$. Since $BD=AB$ , we get $∠BDF=∠BAF= ½∠ABC =∠ABP =∠CBP$. Then line \[FD||PB\]. Hence, $ΔDFE$ is similar to $ΔPBE$. Since $BF||CA$ and $M$ is the midpoint of $AC$, so $E$ is the midpoint of $FB$, i.e. $FE=BE$. Then $ΔDFE$ is congruent to $ΔPBE$. Hence, $FD=PB$. This along with $DB = BA$ and $∠BDF=∠ABP$ imply $ΔBDF$ is congruent to
$Δ ABP$. Therefore, $∠BAP =∠DBF=∠ACB$.