smc

Since $△DEF$ is an equilateral triangle, all of the angles are $60^{\circ }$. The angles in a line add up to $180^{\circ }$, so $$\angle FDB=120-b$$ $$\angle EFC=120-a$$ The angles in a triangle add up to $180^{\circ }$, so $$\angle ABC=60+b-a$$ $$\angle ACB=60-c+a$$ Since $△ABC$ is isosceles and $AB=AC$, by Base-Angle Theorem, $$60+b-a=60-c+a$$ $$b+c=2a$$ $$a=\frac{b+c}{2}$$ The answer is $\boxed{a=\frac{b+c}{2}}$.