2021 AMC 12A Fall Problem 6

Question

As shown in the figure below, point

E

lies on 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

Around point

D,

the triangle’s angle is

360^\circ

minus

\angle ADC+\angle CDE

Triangle

DFE

is isosceles with

DF = DE,

so its two base angles are equal

Because

ABCD

is a square,

\angle ADC = 90^\circ.

Since

E

and

A

lie on opposite sides of line

CD,

ray

DE

is swung past

DC,

so the angle of triangle

DFE

at

D

(with

F

on

\overline{AD}

) is

\angle FDE = 360^\circ

- (\angle ADC + \angle CDE)

= 360^\circ - (90^\circ + 110^\circ)

= 160^\circ.

Since

DF = DE,

triangle

DFE

is isosceles with base angles

\angle DFE = \tfrac{180^\circ - 160^\circ}{2} = 10^\circ.

As

A,

F,

D

are collinear,

\angle AFE = 180^\circ - \angle DFE = 170^\circ.

Thus, the correct answer is D.

Problem 6 in other years