Asking for a hint to prove that altitudes of a triangle meet in a point.

Question

Problem: Use cyclic quadrilaterals to give another proof that altitudes of a triangle meet in a point, as follows. Let $ABC$ be the given triangle. Let the altitudes $BL$ and $CK$ meet at $H$. Let $AH$ meet the opposite side at $M$. Then show that $AM \perp BC$.

Attempt: All that I did is here, I drew $KL$, and found that the two polygons colored in Red and Black are cyclic. I just need a hint to continue on my own.

Answer 1

(I've verified that the angles can be labelled angles can be deduced.)

You are very close. Find more cyclic quads via angle chasing.

Obvious hint (IE This is a natural step when problem solving):
Assuming that $AM$ was truly an altitude, we know that $ALMB$ is a cyclic quad.
So, can you prove that $ALMB$ is a cyclic quad without the assumption?
If yes, we should likely be able to backtrack the steps and conclude that $AM$ is an altitude.

you have $\angle LAM = \angle LBM = \alpha$,

$ $

so $ ALMB$ is a cyclic quad,

$ $

so $\angle AMB = \angle ALB = 90^ \circ $.