Dutch Mathematical Olympiad

a. We first prove the similarity $\triangle CMD \sim \triangle ABC$. Since $BC$ and $MD$ are parallel, we find that $\angle ADM = \angle ACB = 90^\circ$ and also $\angle AMD = \angle ABC$. It follows that $\triangle ABC \sim \triangle AMD$. Because $|AB| = 2|AM|$ we also have that $|AC| = 2|AD|$ and thus $|AD| = |DC|$. This implies the congruence $\triangle AMD \cong \triangle CMD$: both triangles have a right angle at $D$ and the two adjacent sides have the same length. Now we have that $\triangle ABC \sim \triangle AMD \cong \triangle CMD$, and so it holds that $\triangle CMD \sim \triangle ABC$.

Now we will prove that $\triangle CME \sim \triangle ABD$. We already know that $\angle ECM = \angle DCM = \angle CAB = \angle DAB$, and also that
$$\frac{|EC|}{|CM|} = \frac{\frac{1}{2}|DC|}{|CM|} = \frac{\frac{1}{2}|CA|}{|AB|} = \frac{|DA|}{|AB|}$$
This implies that $\triangle CME \sim \triangle ABD$: the triangles have one equal angle and the two adjacent sides have the same ratio.

b. Let $F$ be the intersection of $BD$ and $CM$. Since $BD$ is perpendicular to $CM$ we have that $\angle BFM = 90^\circ$. So in the triangle $\triangle BFM$ we have that $\angle BMF + \angle FBM = 90^\circ$. Because of the similar triangles in part (b) we have $\angle FBM = \angle ABD = \angle CME = \angle FME$. It follows that $\angle BMF + \angle FME = 90^\circ$, hence $EM$ is perpendicular to $AB$. $\square$