Question 176
Let $ABC$ be a right-angled triangle with $\hat{A}=90^{\circ}$. Let $K$ be the midpoint of $BC$, and let $AKLM$ be a parallelogram with centre $C$. Let $T$ be the intersection of the line $AC$ and the perpendicular bisector of $BM$. Let $\omega_{1}$ be the circle with centre $C$ and radius $CA$ and let $\omega_{2}$ be the circle with centre $T$ and radius $TB$. Prove that one of the points of intersection of $\omega_{1}$ and $\omega_{2}$ is on the line $LM$.