Square with midpoints of $AD$ and $CD$ 
Square $ABCD$ is given; $MA=MD=ND=NC$. Show $AF=AB$.
  

The first thing I noticed was $\triangle CDM \cong \triangle BCN$ and we obtain $CM = BN$ and $\angle MCD = \angle NBC$. Now I am trying to show $\angle BFM = 90 ^\circ$ but I am having difficulties with this and don't know if it will help with the solution. Would appreciate help of any kind!
Edit: Thank you for your help! Now I see how $\angle BFM = 90^\circ$. $\angle BAM + \angle BFM = 180 ^\circ$, thus $ABFM$ is a cyclic quadrilateral. Does this help?
 A: Yes, they are perpendicular.  $\angle BNC \cong \angle CMD$ by CPCTC, and therefore $\triangle CFN \sim \triangle CDM$ and thus $\angle CFN$ is a right angle.
Moving on from there, let's just draw parallels to BM and CN through A, B, C, and D.

From this, it should be pretty clear that $AB=AF$.
A: Extend $FM$ and $AB$ to meet at $K$.
Thus, since $\measuredangle KFB=90^{\circ}$ and $FA$ is a median of $\Delta KFB,$ we obtain:
$$KA=AB=FA$$ and we are done!
A: Let $S$ be a center of a square. 
Notice that rotation around $S$ for $90^{\circ}$ takes $BN$ to $CM$ ($B\mapsto C$, $C\mapsto D$ ... and since $DM = CN$ we have $N\mapsto M$) so $\angle BFM = 90^{\circ}$.
Now we see that $ABFM$ is cyclic, so $$<AFB = <AMB = <DMC = 90-<DCM = 90-<CBN = < ABF$$
So $ABF$ is isosceles.
A: Drop perpendicular $AG$ and denote $\angle ABG=x$:
$\hspace{3cm}$
Note:
$$x=\angle ABG=\angle BNC \quad \text{(because $AB||CD$})\\
\angle CFN=90^\circ \quad \text{(because $\Delta CFN\sim \Delta CDM$)} \\
\Delta ABG\cong \Delta BCF \quad \text{(because correspondig angles and one side are equal)}\\
BF=2CF=4FN \quad \text{(because $\tan x=2$)}\\
FG=BF-BG=2CF-BG=2BG-BG=BG.$$
So, the perpendicular $AG$ is also median, hence $\Delta AFB$ is an isosceles triangle and $AF=AB$.
