Let $K$ be midpoint of the hypotenuse of a right triangle $ABC$.On the leg $AB$ is a point $M$ s.t $BM=2MC$.Show that $MAB$ and $MKC$ are similar.

Question

Let $K$ be the midpoint of the hypotenuse of a right triangle $\triangle ABC$.
On the leg $BC$ is a point $M$ such that $ BM = 2MC$ .
Prove that the triangles $\triangle MAB$ and $\triangle MKC$ similar .
My work:
We know that $2*BK=AB$, $2*AK=AB$
If $\triangle MAB$ and $\triangle MKC$ are similar . Then we know $$\frac{MA}{MK}=\frac{AB}{KC}=\frac{MB}{MC}=k$$ I only know this: $\frac{MB}{MC}=\frac{2MC}{MC}=2=k$

Answer 1

We have $\frac{MB}{MC}=2=\frac{BA}{CK}$ (because $K$ is the centre of the circle $ABC$).Also $\angle MBA=\angle MCK$ because triangle $KBC$ is isosceles.