2021 AMC 10A Fall Problem 7

Question

As shown in the figure below, point

E

lies in the opposite half-plane determined by line

CD

from point

A

so that

\angle CDE = 110^\circ.

Point

F

lies on

\overline{AD}

so that

DE=DF,

and

ABCD

is a square. What is the degree measure of

\angle AFE?

160

164

166

170

174

Find the large angle

\angle FDE

around point

D

Since

DE=DF,

triangle

DFE

is isosceles

Since

\angle ADC = 90^{\circ},

we get that

\angle FDE = 360^{\circ} - 90^{\circ} - 110^{\circ}

= 160^{\circ}.

Also since

\triangle FDE

is isosceles, we get that

\angle EFD = \dfrac{180^{\circ} - 160^{\circ}}{2} = 10^{\circ}.

Finally, we get that

\angle AFE = 180^{\circ} - 10^{\circ} = 170^{\circ}.

Thus, D is the correct answer.

Problem 7 in other years