Školjka

Let $ABC$ be a triangle with $|AB| \neq |AC|$. The points $K$, $L$, $M$ are the midpoints of the sides $BC$, $CA$, $AB$, respectively. The inscribed circle of $ABC$ with centre $I$ touches the side $BC$ at point $D$. The line $g$, which passes through the midpoint of segment $ID$ and is perpendicular to $IK$, intersects the line $LM$ at point $P$. \\\\Prove that $|\angle{PIA}| = 90^{\circ}$.