South African Mathematics Olympiad

First note that reflecting the figure around the line $BD$ doesn't change the figure (the equilateral triangle $AEB$ reflects to the triangle $CBF$). This means that $DE$ and $DF$ are symmetric about the line $BD$, which means that $DF=DE$. If we can show that $\angle FDE=60^{\circ }$, then triangle $DEF$ is an isosceles triangle with one $60^{\circ }$ angle, which means it's equilateral.

Now, since $DA=AB=AE$, triangle $ADE$ is isosceles. Next, $$\angle DAE=90^{\circ }-\angle EAB=90^{\circ }-60^{\circ }=30^{\circ }\text{ and so }\angle ADE=\frac{1}{2}(180^{\circ }-\angle DAE)=\frac{1}{2}(180^{\circ }-30^{\circ })=75^{\circ }.$$ Similarly, $\angle DFC=75^{\circ }$.