Let $ABC$ be a triangle. A point $P$ lies on the segment $BC$ such that the circle with diameter $BP$ passes through the incenter of $ABC$. Show that $\frac{BP}{PC}=\frac{c}{s-c}$ where $c$ is the length of segment $AB$ and $2s$ is the perimeter of $ABC$.
As $BP$ is diameter than $PI \perp IB$ where $I$ is incenter.
Let incircle touch $AB$ at $E$. Then $EB=s-b, EI^2=\frac{(s-a)(s-b)(s-c)}{s}$ so $IB^2=EB^2+EI^2= (s-b) (s-b+ \frac{(s-a)(s-c)}{s})=\frac{ac(s-b)}{s}$
$\cos \angle ABC = \frac{a^2+c^2-b^2}{2ac}=2 \cos^2 \frac{\angle ABC}{2}-1$ so $ \cos^2 \frac{\angle ABC}{2} = \frac{s(s-b)}{ac}$
$BP=\sqrt{\frac{IB^2}{ \cos^2 \frac{\angle ABC}{2} }}=\sqrt{\frac{a^2c^2}{s^2}}=\frac{ac}{s}$
$PC=BC-BP=\frac{a(s-c)}{s}$
So $\frac{BP}{PC}=\frac{c}{s-c}$