In triangle $ABC$, where $\angle A=2\angle C$, the internal angle bisector $AM$ is drawn such that $BM = 4$ and $MC = 5$. Let $I$ be the incenter of $\triangle ABC$. The angle bisector $BI$ intersects $AC$ in $P$. Calculate $PI$.
My progress:
$\triangle ABC \sim \triangle MBA\implies \frac{AB}{4}=\frac{9}{AB}\therefore AB = 6\\ \frac{AC}{5}=\frac{AB}{4} \implies AC =\frac{15}{2}\\ \triangle ABC:\frac{AP}{PC} = \frac{AB}{BC}=\frac{6}{9}=\frac{2}{3}\\ AP+PC = \frac{15}{2} \implies \frac{2PC}{3}+PC = \frac{15}{2}\therefore PC = \frac{9}{2}, AP =3\\ \triangle ABP: \frac{AP}{AB}=\frac{PI}{BI}=\frac{3}{6}\implies BI=2PI\\ BP = \frac{2}{b+c}\sqrt{bcp(p-a)} = \frac{2}{9+6}\sqrt{9.6.\frac{45}{4}.(\frac{45}{4}-\frac{15}{2})}=\frac{9\sqrt2}{2} \\ \therefore PI =\frac{3\sqrt2}{2}\approx 2,12 $
