1977 AHSME Problems/Problem 4

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Solution

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, $\measuredangle CDE=\measuredangle BDF=65^\circ$. $\measuredangle EDF=180^\circ - 2(65^\circ)=\textbf{(C) }50^\circ$