Two equal line segments $AB$ and $CD$ intersect each other at a point $M$. If the perpendicular bisectors of $AD$ and $BC$ intersect each other at the point $N$, prove that the two angles $\angle AMN$ and $\angle CMN$ are equal.
According to the following picture, here is the summary of the problem
Theorem. If the following assumptions hold $$\begin{align} AB&=CD \\ NP &\bot AD \\ AP &= PD \\ NQ &\bot BC \\ BQ &=QC \\ \end{align}$$ then $$\angle AMN = \angle CMN$$
My Attempt:
We know that the distance of every point on the perpendicular bisector from two ends of segment is equal. From that, we can know $BN = CN$ and $AN = DN$. But how can we now prove that $MN$ is the bisector of $\angle AMC$?
Since same colored lines are equal in pairs, $\triangle DNC$ is congruent to $\triangle ANB$.
Result-1 $\gamma$ will be equal to $\delta$. From which, we get $\angle DAN = \angle BCN$.
Result-2 $\alpha = \beta$. This means $ADMN$ is cyclic and therefore $x = \angle DAN$.
Hope you can finish the missing detail.