In this diagram $AB$ and $AC$ are the equal sides of an isosceles $\triangle ABC$, in which is inscribed equilateral $\triangle DEF$. 
Designate $\angle BFD$ by $a$, $\angle ADE$ by $b$, and $\angle FEC$ by $c$. Then:

$\textbf{(A)}\ b=\frac{a+c}{2}\qquad \textbf{(B)}\ b=\frac{a-c}{2}\qquad \textbf{(C)}\ a=\frac{b-c}{2} \qquad \textbf{(D)}\ a=\frac{b+c}{2}\qquad \textbf{(E)}\ \text{none of these}$
Since $\triangle DEF$ is an [equilateral triangle](https://artofproblemsolving.com/wiki/index.php/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 $\triangle ABC$ is [isosceles](https://artofproblemsolving.com/wiki/index.php/Isosceles_triangle) and $AB = AC$, by Base-Angle Theorem,
\[60 + b - a = 60 - c + a\]
\[b + c = 2a\]
\[a = \frac{b+c}{2}\]
The answer is $\boxed{\textbf{(D)}}$.