HMMT February, 2018

Solution:
Let $\angle ABD = \angle ADB = x$, and $\angle DAB = 180^{\circ} - 2x$. In triangle $ACD$, by the law of sines,
$$CD = \frac{AD}{\sin \angle ACM} \cdot \sin(198^{\circ} - 2x),$$
and by the law of sines in triangle $BCD$,
$$CD = \frac{BD}{\sin \angle BCM} \cdot \sin(x + 36^{\circ}).$$
Combining the two, we have
$$2 \cos x = \frac{BD}{AD} = \frac{\sin(198^{\circ} - 2x)}{\sin(x + 36^{\circ})} \cdot \frac{\sin \angle BCM}{\sin \angle ACM}.$$
But by the ratio lemma,
$$1 = \frac{MB}{MA} = \frac{CB}{CA} \cdot \frac{\sin \angle BCM}{\sin \angle ACM},$$
meaning that
$$\frac{\sin \angle BCM}{\sin \angle ACM} = \frac{CA}{CB} = \frac{\sin 36^{\circ}}{\sin 18^{\circ}} = 2 \cos 18^{\circ}.$$
Plugging this in and simplifying, we have
$$2 \cos x = \frac{\sin(198^{\circ} - 2x)}{\sin(x + 36^{\circ})} \cdot 2 \cos 18^{\circ} = \frac{\cos(108^{\circ} - 2x)}{\cos(54^{\circ} - x)} \cdot 2 \cos 18^{\circ},$$
so that
$$\frac{\cos x}{\cos 18^{\circ}} = \frac{\cos(108^{\circ} - 2x)}{\cos(54^{\circ} - x)}.$$
We see that $x = 36^{\circ}$ is a solution to this equation, and by carefully making rough sketches of both functions, we can convince ourselves that this is the only solution where $x$ is between $0$ and $90$ degrees. Therefore $\angle ABD = \angle ADB = 36^{\circ}$, $\angle DAB = 108^{\circ}$.

Simple angle chasing yields $\angle AEB = 72^{\circ}$, $\angle ECA = 54^{\circ}$, $\angle EAC = 54^{\circ}$, $\angle EAB = 72^{\circ}$, making $D$, $A$, and $E$ collinear, and so $\angle BDE = 36^{\circ}$. And because $AF = AB = AD$, $\angle FDB = \frac{1}{2} \angle FAB = 9^{\circ}$, so $\angle FDE = 36^{\circ} - 9^{\circ} = 27^{\circ}$.