South African Mathematical Olympiad Third Round

We denote the angles of the triangle by $\alpha = \angle BAC$, $\beta = \angle ABC$ and $\gamma = \angle ACB$. $X$ is the intersection of $BC$ with the angle bisector of $\angle BAC$. Since $\angle BDA = \angle BEA = 90^\circ$, both $D$ and $E$ lie on the circle with diameter $AB$. Thus $AEBD$ is cyclic, which implies that
$$\angle EDB = \angle EAB = 180^\circ - \alpha.$$
It is given that $DE$ and $CF$ are parallel, hence
$$\angle EDB = \angle FCB = 90^\circ - \angle FBC = 90^\circ - \beta,$$
so $\beta = \alpha - 90^\circ$. Likewise,
$$\angle FDC = \angle FAC = 180^\circ - \alpha,$$
and
$$\angle FDC = \angle AXC = 180^\circ - \angle ACX - \angle XAC = 180^\circ - \gamma - \alpha/2,$$
so $\gamma = \alpha/2$. Since $\alpha + \beta + \gamma = 180^\circ$, this gives us
$$\alpha + (\alpha - 90^\circ) + \alpha/2 = 180^\circ,$$
and thus $\alpha = 108^\circ$, $\beta = 18^\circ$ and $\gamma = 54^\circ$.