Geometry Question: proving that NM and MC are perpendicular I have been stuck on this problem for quite a while:

Let ABCD be a rectangle and BD its diagonal. Take CE$\perp$BD and M the midpoint of DE. Then, if N is the midpoint of AB, prove that $\angle{NMC}=90^{\circ}$

I have tried pretty much everything, creating other rectangles inside the rectangle but I can't get how to use the fact that M is the midpoint of DE. I also tried going about this by naming every single angle and taking the relationships between them, but no luck so far. Here is the shape I made in Geogebra. I can solve it using analytic geometry, but I am looking for a purely (euclidean) geometrical solution. Thanks in advance! 
 A: General answer, if $BN/BA=EM/ED=r$
Suppose $AB=l$ and $BC=w$. Let $\angle BNC=\alpha$ and $\angle CBD=\beta$. We want to prove that $\angle CNM=\beta$, so that we can conclude that because $\angle CNM=\angle CBD$, $BCMN$ is a cyclic quadrilateral.
Consider that $BE/BC=AD/BD$, so
$$BE=BC(AD/BD)=\frac{w^2}{\sqrt{l^2+w^2}}$$
and
$$ED=BD-BE=\sqrt{l^2+w^2}-\frac{w^2}{\sqrt{l^2+w^2}}$$
With the Law of Sines, we can obtain
$$\frac{BN}{\sin\angle BMN}=\frac{BM}{\sin\angle BNM}$$
$$\frac{rl}{\sin\angle BMN}=\frac{BE+rED}{\sin(90^{\circ}+\beta-\angle BMN)}$$
$$\frac{\sin(90^{\circ}+\beta-\angle BMN)}{\sin\angle BMN}=\frac{BE+rED}{rl}$$
$$\frac{\cos(\angle BMN-\beta)}{\sin\angle BMN}=\frac{r\sqrt{l^2+w^2}+(1-r)w^2/\sqrt{l^2+w^2}}{rl}$$
$$\frac{\cos\angle BMN\cos\beta+\sin\angle BMN\sin\beta}{\sin\angle BMN}=\frac{r\sqrt{l^2+w^2}+(1-r)w^2/\sqrt{l^2+w^2}}{rl}$$
$$\cot\angle BMN\cos\beta+\sin\beta=\frac{r\sqrt{l^2+w^2}+(1-r)w^2/\sqrt{l^2+w^2}}{rl}$$
$$\cot\angle BMN\left(\frac{w}{\sqrt{l^2+w^2}}\right)+\frac{l}{\sqrt{l^2+w^2}}=\frac{l^2+w^2}{l\sqrt{l^2+w^2}}+\frac{\frac{1-r}{r}w^2}{l\sqrt{l^2+w^2}}$$
$$\cot\angle BMN\left(\frac{w}{\sqrt{l^2+w^2}}\right)=\frac{w^2}{l\sqrt{l^2+w^2}}+\frac{\frac{1-r}{r}w^2}{l\sqrt{l^2+w^2}}$$
$$\cot\angle BMN\left(\frac{w}{\sqrt{l^2+w^2}}\right)=\frac{w^2/r}{l\sqrt{l^2+w^2}}$$
$$\cot\angle BMN=\frac{w}{rl}$$
$$\tan\angle BMN=\frac{rl}{w}=\tan(90^{\circ}-\alpha)$$
Because $\angle BMN$ is acute, $\angle BMN=90^{\circ}-\alpha$. Therefore
$$\angle CNM=180^{\circ}-\angle ABD-\angle BNC-\angle BMN$$
$$\angle CNM=180^{\circ}-90^{\circ}+\beta-\alpha-90^{\circ}+\alpha$$
$$\angle CNM=\beta$$
Finally, we conclude that because $\angle CNM=\angle CBD$, $BCMN$ is a cyclic quadrilateral, and therefore
$$\angle NMC = 180^{\circ}-90^{\circ}$$
$$\angle NMC=90^{\circ}$$
Q.E.D.
A: Here is a geometric proof.

We have $\triangle ABC \sim \triangle DEC$. Thus
$$\frac{AB}{BC}=\frac{DE}{EC}\Rightarrow \frac{AB/2}{BC}=\frac{DE/2}{EC}\Rightarrow \frac{NB}{BC}=\frac{ME}{EC} \tag{1}$$
In $\triangle$s $NBC$ and $MEC$, $\angle MEC = 90^\circ = \angle NBC$ and from $(1)$,
$$\frac{NB}{ME}=\frac{BC}{EC}$$
Hence the two triangles are similar by $SAS$ criterion. Due to this $\angle CME = \angle CNB$ and the quadrilateral $CMNB$ turns out to be cyclic. It follows then that $\angle CMN = 180^\circ - \angle NBC = 90^\circ. \quad \square$
A: 
From triangles similar to $\triangle BEC$:$$\frac{x}{y} = \frac{w}{z} = \frac{2z}{x}\\x^2=2yz\\2z^2=wx$$
In order to prove that the shaded triangles are similar (which implies $\angle{NMC}=90^{\circ}$) we must show that $$\frac{y+z}{\frac{x}{2}}=\frac{w+x}{z}$$
And this is simple:$$\frac{2(y+z)}{x}=\frac{w+x}{z}$$
Cross multiplication:$$2yz+2z^2=wx+x^2$$
Substitution:$$x^2+wx=wx+x^2$$
QED
