prove $\angle BAP= \angle ACB$

[Tahmid Hasan]

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 $ \angle ABC$ intersect line $DM$ at $P$. Prove that $\angle BAP =\angle ACB$.

[Matheater]

Let us join $A,D$.Since$\angle DAB= \angle ABP=\frac{1}{2} \angle ABC$, it shows $AD \parallel BP$,now let us draw a point $T$ on $AD$ such that $AC \parallel TB$.Let us define $DM \cap TB=X$.Now One can notice that a homothety with center $D$ and ratio $\frac {AD}{DT}$ that sends $M$ to $X$,so $TX=BX$.From here on we conclude $\triangle TDX \cong \triangle BPX (ASA)$.Hence $DT=BP$.So it becomes clear to us that $\triangle DTB \cong \triangle ABP (SAS)$.Then $\angle BAP=\angle TDB=\angle ACB$
my first post here ,sorry if it is wrong.