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 $\angle AFE?$

$160$

$164$

$166$

$170$

$174$

Answer: D

Concepts:angle chasing isosceles triangle square (geometry)

Difficulty rating: 960

Solution:

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.

2021 AMC 10A Fall Problem 7

Answer: D

Difficulty rating: 960

Solution:

Problem 7 in Other Years