Nice and easy problem ! Suppose that AB<AC. $ \angle OAN=\angle OAB+\angle BAN=\angle CAD+\angle BAD=\angle BAC$. Similarly, $ \angle OAM=\angle BAC$, wich imlies AO is the bisector of angle MAN. Denote H,K are the orthogonal projections of E on AN,AM respectively. Because of symmetry, we have some equalities of angles: $ \angle ANB=\angle ADB=90^o\Rightarrow BN\parallel EH\Rightarrow\angle BNE=\angle NEH$ Similarly, $ \angle EMC=\angle MEK$ $ \angle MEK=\angle NEH$ Hence result.

Mine a little different solution.. $ \left\| \begin{array}{cc} \angle{ANB} = \angle{AMC} = 90^\circ \\ \angle{NAB} = 90^\circ - \angle{B} \\ \angle{MAC} = 90^\circ - \angle{C} \\ \angle{BAE} = 90^\circ - \angle{C} \\ \angle{EAC} = 90^\circ - \angle{B} \end{array} \right\| \Rightarrow \angle{NAE} = \angle{EAM} = 180^\circ - \angle{B} - \angle{C} = \angle{A}$ Also $ \left\|\begin{array}{cc} AN = AD \\ AM = AD \end{array} \right\| \Rightarrow AN = AM$ In $ \triangle{NAE}$ and $ \triangle{EAM}$ we have $ \left\|\begin{array}{cc} AN = AM \\ AE \ - \ \ \mathrm{common \ side} \\ \angle{NAE} = \angle{EAM} \end{array} \right\| \Rightarrow \triangle{NAE} \equiv \triangle{EAM} \Rightarrow \angle{ANE} = \angle{AME}$. But $ \left\|\begin{array}{cc} \angle{BNE} = 90^\circ - \angle{ANE} \\ \angle{EMC} = 90^\circ - \angle{AME} \end{array} \right\| \Rightarrow \boxed{\angle{BNE} = \angle{EMC}}$

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