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 PIC = 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, in this case, $ \angle APC$ if the angle bisector of $C$ intersects $MK$ at $P$. So it´s like I know how to prove the original theorem but not its converse. Can you help me please?