The 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 $\gamma =54^{\circ }$.