Point $\displaystyle M$ is the middle of side $\displaystyle BC$ in acute triangle $\displaystyle \Delta ABC$. Point $\displaystyle K$ lies on side $\displaystyle BC$ in such a way that $\displaystyle \angle BAM = \angle KAC$. A point $\displaystyle E$ was chosen on side $\displaystyle AK$ such that $\displaystyle \angle BEK = \angle BAC$/ Show that $\displaystyle \angle KEC = \angle BAC $.

I ask for your kind help in tackling this. I would attach a picture, but I will not, as I am not completely sure how to construct the given points so that it all hangs together.