Point M is the middle of a BC side of an acute triangle ABC. Point K lies on the BC side and fills the condition $\displaystyle \angle BAM=\angle KAC$. On the AK segment such point E was chosen that $\displaystyle \angle BEK=\angle BAC$. Prove that $\displaystyle \angle KEC = \angle BAC$.

From what see in my provisory image, the M point is the same point as the K one since KAC divides the BAC angle in the half and thus for BAM to be the other half, M must be K. Then, we choose an E on AK and since we know that AK (or AM, it's the same) divides BAC in the half, automatically BEK = KEC and since we've selected such E that BEK=BAC, KEC=BAC.

Is my thinking right? Cause frankly - I don't think so..