1977 AHSME Problems/Problem 4

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

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