Let $\Delta ABC, \angle BCA=90^o$ draw $CH$ is perpendicular to $AB$ at $H$, the circle $A$ with AC is radius and the circle B with BC is radius. $I$ is chosen arbitrarily in $CH$. $AI\cap (B,BC)={K,Q} (\text {K}\in AI)$ and $BI\cap (A,AC)={L,P} (\text {L}\in BI)$ and $BK\cap AL={M}$ . Prove that $\Delta KML$ is isosceles triangle.
I proved that $PKLQ$ is cyclic quadrilateral
Then i will prove $\angle LKM=\angle MLK$
Or i will prove it by trigonometric $\sin \angle LKM=\sin \angle MLK$ but failed.