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 $\angle AFE?$

$160$

$164$

$166$

$170$

$174$

Solution:

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.

2021 AMC 12A Fall Problem 6

Solution:

Problem 6 in Other Years