CH isn't well defined

there is only one height of trapezoid passing through $C$, where $CD$ is one base, meaning that $H$ lies on other line of base $AB$

@above No, $AB$ can't be base, because $AB = CD$ then means that $ABCD$ - parallelogram and $BC=AD$. Contradiction. So it must be isoscale trapezoid with bases $AD, BC$ and $H \in AD$.

Attachments:

I was reading $AB//CD$ instead of the correct $AB=CD$, (indeed it is an isosceles trapezoid)

Solution: Let $AM \cap CK = P$. Then $P$ - centroid of $\triangle ACD$ $\implies CP/PK=2$. Also $BC/KH=2$. So, $P$ - center of homothety sending $KH$ to $CB$.