1977 AMC 12 Problem 4

Attempt Problem 4 of the 1977 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1977 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

In triangle

ABC,

AB=AC

and

\angle A=80^\circ.

If points

D,

E

and

F

lie on sides

BC,

AC

and

AB,

respectively, and

CE=CD

and

BF=BD,

then

\angle EDF

equals

30^\circ

40^\circ

50^\circ

65^\circ

none of these

First determine the two base angles of

\triangle ABC

Use the two given equal-length pairs to find

\angle EDC

and

\angle BDF

The base angles of

\triangle ABC

are each

50^\circ.

Since

CE=CD,

triangle

CED

has vertex angle

50^\circ,

\angle EDC=65^\circ.

Similarly,

BF=BD

gives

\angle BDF=65^\circ.

The three angles above the straight line

CB

sum to

180^\circ,

\begin{aligned} \angle EDF&=180^\circ-65^\circ-65^\circ\\ &=50^\circ. \end{aligned}

Therefore, the correct answer is C.

Problem 4 in other years