1977 AHSME Problems/Problem 4

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Problem 4

[asy] size(130); pair A = (2, 2.4), C = (0, 0), B = (4.3, 0),  E = 0.7*A, F = 0.57*A + 0.43*B, D = (2.4, 0); draw(A--B--C--cycle); draw(E--D--F); label("$A$", A, N); label("$B$", B, E); label("$C$", C, W); label("$D$", D, S); label("$E$", E, NW); label("$F$", F, NE); //Credit to MSTang for the diagram [/asy]

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

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

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