# 2021 Fall AMC 12A Problems/Problem 6

The following problem is from both the 2021 Fall AMC 10A #7 and 2021 Fall AMC 12A #6, so both problems redirect to this page.

## Problem

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 $\angle AFE$? $[asy] size(6cm); pair A = (0,10); label("A", A, N); pair B = (0,0); label("B", B, S); pair C = (10,0); label("C", C, S); pair D = (10,10); label("D", D, SW); pair EE = (15,11.8); label("E", EE, N); pair F = (3,10); label("F", F, N); filldraw(D--arc(D,2.5,270,380)--cycle,lightgray); dot(A^^B^^C^^D^^EE^^F); draw(A--B--C--D--cycle); draw(D--EE--F--cycle); label("110^\circ", (15,9), SW); [/asy]$ $\textbf{(A) }160\qquad\textbf{(B) }164\qquad\textbf{(C) }166\qquad\textbf{(D) }170\qquad\textbf{(E) }174$

## Solution

By angle subtraction, we have $\angle ADE = 360^\circ - \angle ADC - \angle CDE = 160^\circ.$ Note that $\triangle DEF$ is isosceles, so $\angle DFE = \frac{180^\circ - \angle ADE}{2}=10^\circ.$ Finally, we get $\angle AFE = 180^\circ - \angle DFE = \boxed{\textbf{(D) }170}$ degrees.

~MRENTHUSIASM ~Aops-g5-gethsemanea2

## Video Solution by TheBeautyofMath

for AMC 10: https://youtu.be/ycRZHCOKTVk?t=232

for AMC 12: https://youtu.be/wlDlByKI7A8

~IceMatrix

## Video Solution by WhyMath

~savannahsolver

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 