Geometry Question: proving that NM and MC are perpendicular

Solution 1:

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

Solution 2:

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.