Given a $\displaystyle \triangle ABC$ and $\displaystyle D$ a point on the extension of side $\displaystyle BC$ (from $\displaystyle B$) such that $\displaystyle \overline{BD}=\overline{BA}$ and let $\displaystyle M$ be the mid point of $\displaystyle \overline{AC}$. The angle bisector of $\displaystyle \measuredangle\,ABC$ intersects to $\displaystyle \overline{DM}$ in $\displaystyle P$. Prove that $\displaystyle \measuredangle\,BAP=\measuredangle\,ACB.$
• Let $\displaystyle N$ be the foot of the angle bisector of $\displaystyle B$ on the side $\displaystyle AC$. Since $\displaystyle \overline{AD}\parallel\overline{BN}\implies\triang le ADM\sim\triangle NPM$, it yields $\displaystyle \frac{{\overline {AD} }} {{\overline {NP} }} = \frac{{\overline {AM} }} {{\overline {MN} }}$
• On the other hand, $\displaystyle \triangle BCN\sim\triangle DCA$, then
$\displaystyle \frac{{\overline {BN} }} {{\overline {AD} }} = \frac{{\overline {CN} }} {{2 \cdot \overline {AM} }} \implies \frac{{2 \cdot \overline {MN} }} {{\overline {NP} }} = \frac{{\overline {CN} }} {{\overline {BN} }} \implies \frac{{\overline {AN} }} {{\overline {CN} }} = \frac{{\overline {BP} }} {{\overline {BN} }}$
• Since $\displaystyle \frac{{\overline {AN} }} {{\overline {CN} }} = \frac{{\overline {AB} }} {{\overline {BC} }} \implies \frac{{\overline {BP} }} {{\overline {BN} }} = \frac{{\overline {AB} }} {{\overline {BC} }}$
We have $\displaystyle \measuredangle\,ABN=\measuredangle\,CBN\implies\tr iangle{ABP}\sim\triangle{CBN}\,\therefore\,\measur edangle\,BAP=\measuredangle\,ACB$