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$$ABC$$ and $$ADE$$ are isosceles triangles. If $$\angle BFD=156^\circ$$, then $$\angle A$$ equals
Let $$\angle A=x$$. Since $$AB=BC$$, the base angle $$\angle ACB=x$$, and since $$AD=DE$$, the base angle $$\angle AED=x$$. In quadrilateral $$ACEF$$, the four interior angles are $$x,x,x$$ and $$156^\circ$$, so $$3x+156^\circ=360^\circ$$ and $$x=68^\circ$$.
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