South African Mathematics Olympiad

We will compute the angle sizes of quadrilateral $DEGB$ and use the fact that the sum of the interior angles of a quadrilateral is equal to $360^{\circ }$.

Since $AFC$ is an equilateral triangle, $\angle FAC=60^{\circ }$ and so $\angle EAF=90^{\circ }-60^{\circ }=30^{\circ }$, which yields $\angle BAE=90^{\circ }+30^{\circ }=120^{\circ }$. Since $AB=AC=AF$, triangle $BAF$ is isosceles, and so $$\angle GBC=\angle FBA=\frac{1}{2}(180^{\circ }-\angle BAF)=\frac{1}{2}(180^{\circ }-120^{\circ })=30^{\circ }.$$ Next, $AE=AB=AC$, and so triangle $EAC$ is a right-angled isosceles triangle, which means that $\angle BCG=\angle ACE=45^{\circ }$. Hence, $$ \angle BGE = \angle GBC + \angle GCB = 30^\circ + 45^\circ = 75^\circ.