SELECTION TESTS OF THE BELARUSIAN TEAM TO THE IMO

Let us rotate $△AED$ around the point $A$ so that $E\to B$, $D\to T$, and reflect $△ABT$ 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

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

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.