I saw this question from a student's geometry homework. And I was thinking whether there was an efficient method to solve the question.
We are supposed to prove $DM \perp MN$.
My thinking is to draw the midpoint of AD and label it as L on AD. Next, connect LM. I was trying to use the midpoint theorem.
We know that $LM \parallel DK$. Since $DK \perp AC$, we have $LM \perp AC$. Therefore, I just need to prove $\angle LMD = \angle NMC$. But I failed to do so.
Does anyone have a better way to prove this question?
Thanks!
Notice that $L$, $D$, $N$, and $C$ are concylic, since they form a rectangle.
Since you showed that $LM\perp MC$, it follows that $\angle LMC$ and $\angle LDC$ are both right angles and hence supplementary, so $M$, $L$, $D$, and $C$ are concyclic as well.
Thus, $M$, $L$, $D$, $N$, and $C$ are all concyclic. Since $DN$ is also a diameter of this circle, it follows that $\angle DMN$ must be a right angle.