BdMO Online Forum

Let us draw the bisectors of $\angle MND $ and $\angle MNC$ which intersects AB at X and Y respectively.
we get the followings:
$\angle ADN=\angle YNC= \angle MNY =\angle MYN$[Using AB||CD(YN transversal) and what is given]
$\angle BCN=\angle XND = \angle MNX = \angle MXN$[Same reason as before]
Now, $\triangle MNX$ and $\triangle MNY$ are isosceles triangles.

$\therefore$ $XM= MN$ and $MY=YN$
$\therefore XM= YN$
And we know, $AM= MB$
$\therefore AY=BX \cdots (1)$


Again ,
$\angle ADN=\angle YNC$ $\therefore AD||NY$ and $AY||DN(Given)$
$\therefore$ $ADNY$ is a parallelogram.
$\therefore AY=DN $
Similarly, $BCNX$ is a parallelogram.
$\therefore BX=NC$
From (1) we get $AY=BX$
$\therefore DN=CN$
$\therefore$ N is the midpoint of CD[proved]