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funni Clearly, $ABCD$ is cyclic. Claim: $AC \perp DE$. Proof. Let $AC$ and $DE$ meet at $F$. We have $$\angle DCA = 90^{\circ}-\angle ACB = 90-\angle ADB = 90 - \angle EDC \rightarrow \angle DCA + \angle EDC = 90 \rightarrow \angle DFC = 90^{\circ},$$as needed. $\square$ Define $O$ as the circumcenter of $\triangle ABC$. More angle chasing gives us $MO \perp AC$ so that $MO$ meets $AC$ at its midpoint, which implies it is the perpendicular bisector of $AC$, and our result follows. $\blacksquare$