1. ## Triangle

Given a $\triangle ABC$ and $D$ a point on the extension of side $BC$ (from $B$) such that $\overline{BD}=\overline{BA}$ and let $M$ be the mid point of $\overline{AC}$. The angle bisector of $\measuredangle\,ABC$ intersects to $\overline{DM}$ in $P$. Prove that $\measuredangle\,BAP=\measuredangle\,ACB.$

• Let $N$ be the foot of the angle bisector of $B$ on the side $AC$. Since $\overline{AD}\parallel\overline{BN}\implies\triang le ADM\sim\triangle NPM$, it yields $\frac{{\overline {AD} }}
{{\overline {NP} }} = \frac{{\overline {AM} }}
{{\overline {MN} }}$
• On the other hand, $\triangle BCN\sim\triangle DCA$, then
$\frac{{\overline {BN} }}
{{\overline {AD} }} = \frac{{\overline {CN} }}
{{2 \cdot \overline {AM} }} \implies \frac{{2 \cdot \overline {MN} }}
{{\overline {NP} }} = \frac{{\overline {CN} }}
{{\overline {BN} }} \implies \frac{{\overline {AN} }}
{{\overline {CN} }} = \frac{{\overline {BP} }}
{{\overline {BN} }}$

• Since $\frac{{\overline {AN} }}
{{\overline {CN} }} = \frac{{\overline {AB} }}
{{\overline {BC} }} \implies \frac{{\overline {BP} }}
{{\overline {BN} }} = \frac{{\overline {AB} }}
{{\overline {BC} }}$
We have $\measuredangle\,ABN=\measuredangle\,CBN\implies\tr iangle{ABP}\sim\triangle{CBN}\,\therefore\,\measur edangle\,BAP=\measuredangle\,ACB$