Final Round, September 2019

Let $I$ be the reflection of point $M$ in the line $AB$. We define $\alpha =\angle CAI$ and $\beta =\angle CBI$. Since $AI$ is the angular bisector of $\angle CAB$, we find that $\angle IAB=\alpha$. Since $I$ is the reflection of $M$ in the line $AB$, we find that $\angle BAM=\alpha$. Triangle $AMC$ is isosceles with apex $M$, because $|AM|=|CM|$. Hence, we see that $\angle MCA=\angle CAM=3\alpha$. In the same way, we see that $\angle IBA=\angle ABM=\beta$ and $\angle MCB=3\beta$. The sum of the angles of triangle $ABC$ is therefore $2\alpha +(3\alpha +3\beta )+2\beta =180^{\circ }$. From this, we conclude that $\alpha +\beta =\frac{180^{\circ }}{5}=36^{\circ }$, and hence that $\angle ACB=3\alpha +3\beta =3\cdot 36^{\circ }=108^{\circ }$.



Since $MAB$ is an isosceles triangle (as $|AM|=|BM|$), we see that $\alpha =\beta =18^{\circ }$. It follows from this that $\angle CAB=2\alpha =\angle ABC$ and therefore that triangle $ACB$ is isosceles. By considering the sum of the angles in triangle $AMC$, we find that $\angle AMC=180^{\circ }-6\alpha =72^{\circ }$. Hence we also find that $\angle CMD=180^{\circ }-\angle AMC=108^{\circ }$. We have already seen that $\angle ACB=108^{\circ }$. It follows that triangles $ACB$ and $CMD$ are both isosceles triangles with an angle of $108^{\circ }$ at the apex. Hence, they are similar triangles. This implies that $\frac{|CM|}{|CD|}=\frac{|AC|}{|AB|}$. By multiplying by both denominators and observing that $|CM|=|AM|$, we obtain the required result.