Dutch Mathematical Olympiad

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

Now we will prove that $△CME\sim △ABD$. 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 $△CME\sim △ABD$: 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 $△BFM$ 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 $AB$. $\square$