I have not a solution yet, so I made a figure !

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The sketch of solution. We need 3 steps to compute angles $ \angle MHN$and $ \angle MCK$. 1. $ CD=CN$. This can be done by a little computing. 2. $ AM^2=AD^2=AH*AC$ so $ \angle BMC=\angle MHC$. 3. $ CN^2=CD^2=CH*CA$ so $ \angle CHN=\angle CNA$. Finally $ \angle MHN=\angle MCK$

This is superb, rchlch!!

Hello! The proposed problem can be generalized: Let $ ABCD$ be a cyclic quadrilateral ($ AD < AB$ and $ \angle ADC = \alpha\leq 90^{\circ}$), $ M\in\left(AB\right)$ so that $ AM = AD$ and the points $ E,F\in AC$ so that $ \angle DEA = \angle DFC = \alpha$. The point $ N\in BC$ so that $ \angle MDN = 90^{\circ} - \alpha$ ($ \angle MDN$ and $ \angle EDF$ have the same orientation) and the points $ X,K\in AN$ so that $ \angle AXC = \alpha$ and $ \angle XCK = \angle ENF$ ($ \angle XCK$ and $ \angle ENF$ have the same orientation). Prove that $ \angle MEN = \angle MCK$.

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Hi Petry, Nice generalization. Same approach. From $ \angle MDN=90^{\circ}-\alpha$, we know $ CD=CN$. Then use the two similar triangles mentioned in rchlch's solution and some angle chasing to get the result.

first,we can get $AD^2=AM^2=AH*AC$; $\angle MDC=\angle MNC=\frac{A}{2}$ hence $DC=CN$ so $DC^2=CN^2=CH*CA$ then$\angle NHM=\angle NHC-\angle MHC=\angle CNA-\angle CMB=90-\angle KCN-\angle CMB=\angle KCM$ QED