In the adjoining figure, triangle $$BCD$$ is equilateral. If $$\angle AFB=90^\circ$$ and $$AH$$ bisects $$\angle FAE$$, then the measure of $$\angle HGE$$ in degrees is
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Since triangle $$BCD$$ is equilateral, the corresponding angle at $$B$$ is $$60^\circ$$. The exterior angle at $$A$$ gives $$\angle BAE=80^\circ$$, while right triangle $$ABF$$ gives $$\angle BAF=30^\circ$$. Hence $$\angle FAE=50^\circ$$, and its bisector makes $$25^\circ$$ with $$AF$$. Therefore, the angle at $$G$$ is $$180^\circ-90^\circ-25^\circ=65^\circ$$, and the vertically opposite angle $$\angle HGE$$ is also $$65^\circ$$.