[1] Triangles BCL and B*KC* are homothetic. Measuring the angle at K gives:

chi + omega = A + pi/2.

angle(B*KC*) = angle( BLC) = pi/2 + pi - (chi+omega) = pi - A.

Since also angle(BHC) = pi - A, {B,H,L,C} are on a circle c'.

[2] By well known property of the orthocenter H, c' coincides with the reflected on BC of the circumcircle of triangle ABC.

[3] This implies that the intersection point L' of AM with c' is the symmetric of A with respect to M and the intersection point of AH with c' is the reflected point of A with respect to BC. Thus angle(HH'L') is a right one and since HKL'H' is cyclic angle(HKL'), opposite to H', is also a right one.

Triangle L

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