In $$\triangle ABC$$ shown below, $$AB=AC$$, $$F$$ is a point on $$AB$$ and $$E$$ a point on $$AC$$ such that $$AF=EF$$, $$H$$ is a point in the interior of $$\triangle ABC$$, $$D$$ is a point on $$BC$$ and $$G$$ is a point on $$AB$$ such that $$EH=CH=DH=GH=DG=BG$$. Also, $$\angle CHE=\angle HGF$$. The measure of $$\angle BAC$$ in degrees is

Let $$\angle BAC=x$$. Since $$AB=AC$$, the base angles give $$\angle1=\angle2=90^\circ-\frac{x}{2}$$. The equal-length conditions create the $$60^\circ$$ angles shown in the construction, giving $$\angle3=120^\circ-x$$ and $$\angle5=\angle4=30^\circ+\frac{x}{2}$$. Using $$\angle7+\angle5=\angle1$$ gives $$30^\circ+\frac{x}{2}+30^\circ+\frac{x}{2}=90^\circ-\frac{x}{2}$$, hence $$x=20^\circ$$.