Selection Tests of the Belarusian Team to the IMO

Let us rotate $\triangle AED$ around the point $A$ so that $E \to B$, $D \to T$, and reflect $\triangle ABT$ symmetrically with respect to $BT$ ($A \to F$). Since $\angle ADE = 150^\circ$, then the triangle $ATF$ is equilateral. Let us prove that $FD = DC$. Denote $\angle EAB = \alpha$ and $\angle AED = \beta$. Then

$$\angle ABC = \angle ABE + \angle CBE = \frac{180^\circ - \alpha}{2} + (60^\circ - \alpha) = 150^\circ - \frac{3\alpha}{2} \text{ and } \angle BAC = 15^\circ + \frac{3\alpha}{4}.$$
$$\begin{aligned} \angle FCD &= \angle FCA - \angle DCA = \frac{1}{2}\angle FBA - (\angle DEA - \angle EAC) = \\ &= \beta - (\beta - \angle EAC) = \angle EAC = \angle BAC - \alpha = 15^\circ - \frac{\alpha}{4}. \end{aligned}$$

Here we used the fact that the points $D$ and $A$ lie in the same half-plane relative to the line $FC$, since $\angle ACD < \angle AED = \beta = \angle ACF$.

Note that $\angle TFD = \frac{1}{2}\angle TAD = \frac{\alpha}{2}$, so
$$\angle DFC = \angle AFC - \angle AFD = \frac{1}{2}\angle ABC - (\angle AFT - \angle TFD) = \left(75^\circ - \frac{3\alpha}{4}\right) - \left(60^\circ - \frac{\alpha}{2}\right) = 15^\circ - \frac{\alpha}{4}.$$

Therefore, $\angle DFC = \angle FCD$, i.e. $FD = DC$. Hence $FC \perp BD$, i.e. $\angle BDC + \angle FCD = 90^\circ$. Recalling that $\angle FCD = \angle EAC$, we obtain the statement of the problem.