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.