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!

## Best Answer

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 $\angle CMN = 180^\circ - \angle NBC = 90^\circ. \quad \square$