Alternative Solution by PSC. Let $P$ be the point of intersection of $E M$ with $A C$. The triangles $A B C$ and $M P C$ are equal since they have equal angles and $M C=\frac{B C}{2}=A C$. They also share the angle $\hat{C}$, so they must have identical incenter.

Let $I$ be the midpoint of $E K$. We have $\angle P E I=\angle B E M=75^{\circ}=\angle E K P$. So the triangle $P E K$ is isosceles and therefore $P I$ is a bisector of $\angle C P M$. So the incenter of $M P C$ belongs on $P I$. Since it shares the same incenter with $A B C$, then $I$ is the common incenter. We can now finish the proof as in the first solution.

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