In triangle $ABC, AB=AC$ and $\measuredangle A=80^\circ$. If points $D, E$, and $F$ lie on sides $BC, AC$ and $AB$, respectively, and $CE=CD$ and $BF=BD$, then $\measuredangle EDF$ equals
$\textbf{(A) }30^\circ\qquad \textbf{(B) }40^\circ\qquad \textbf{(C) }50^\circ\qquad \textbf{(D) }65^\circ\qquad \textbf{(E) }\text{none of these}$
Solution by e_power_pi_times_i
Because $\measuredangle A=80^\circ$, $\measuredangle B=\measuredangle C=50^\circ$. Then, because triangles $CDE$ and $BDF$ are isosceles, $\measuredangle CDE=\measuredangle BDF=65^\circ$. $\measuredangle EDF=180^\circ - 2(65^\circ)=\textbf{(C) }50^\circ$