# Geometry Question: proving that NM and MC are perpendicular

euclidean-geometrygeometry

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!

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