The incircle of $ABC$ touches $BC$, $CA$, $AB$ at $K$, $L$, $M$ respectively. The line through $A$ parallel to $LK$ meets $MK$ at $P$. Show that $ \angle API = 90$ and that points $P,I$ and $C$ are collinear. $I$ is the incenter of $\triangle ABC$.
It´s easy to show that $ \angle API = 90$. $ \angle APM + \angle MKL = 180$ but $\angle MKL = \angle ALM$ then $ \angle APM + \angle ALM = 180$ and $APLM$ would be cyclic. Now $ \angle API + \angle ALI = 180$, since angle $\angle ALI =90$ then it follows that $ \angle API = 90$. But I do not how to show the other part.
I know that there is a theorem that involves a touch chord, a midline and an angle bisector, and a consequence of that theorem is that if you call, in this case, $P$ intersection of the angle bisector (I say $P$ is in the angle bisector since that´s what they are asking us to show) and a touch chord then $ \angle API = 90$ ($A$ would be a vertex of the triangle, I think it has to be collinear with the point where the angle bisector begins and the missing touch point). So it´s like I know how to prove the original theorem but not its converse. Can you help me show that $P,I$ and $C$ are collinear ?
Let the point $X$ be the intersection of lines $AP$ and $CB$. Since $\triangle{CLK}$ is isosceles and $AX \parallel LK$, $\triangle{CAX}$ is isosceles. The line $CI$ is then perpendicular to line $AX$, and since $AP\perp PI$ it follows that $P$ is on line $CI$ (because there is only one line going through $I$ that is perpendicular to $AX$).