All-Soviet-Union Mathematical Olympiad

Solution:
The idea is to find an expression for the perpendicular distance $h$ from $D$ to $AB$. Let $\gamma = \angle ACB$, $\alpha = \frac{1}{2} \angle CAB$, and $\beta = \frac{1}{2} \angle ABC$. We have $h = AP \sin \alpha$.

Using the sine rule on $APC$, we have $AP = AC \frac{\sin (\gamma + \beta)}{\sin (\alpha + \beta)}$, so $h = AC \frac{\sin \alpha \sin (\gamma + \beta)}{\sin (\alpha + \beta)}$. Similarly, the perpendicular distance $k$ from $E$ to $AB$ is $BC \frac{\sin \beta \sin (\gamma + \alpha)}{\sin (\alpha + \beta)}$.

We also have that $\frac{AC}{BC} = \frac{\sin 2\beta}{\sin 2\alpha}$, and hence $\frac{h}{k} = \frac{\sin 2\beta \sin \alpha \sin (\gamma + \beta)}{\sin 2\alpha \sin \beta \sin (\gamma + \alpha)}$. Using the fact that $\sin (\gamma + \beta) = \sin (2\alpha + \beta)$, and the expression for $\sin 2\theta$, we get $\frac{h}{k} = \frac{\sin (2\alpha + 2\beta) + \sin 2\alpha}{\sin (2\alpha + 2\beta) + \sin 2\beta}$ and hence $h = k$ iff the triangle is isosceles.

For some reason the geometric solution took me longer to find. Let $ED$ meet $BC$ at $X$. Then $XCD$ and $XBE$ are isosceles, so $BC = BX + XC = DX + XE = DE$. Similarly, $AC = DE$. Hence $AC = BC$.