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2021 Fall AMC 10A Problems/Problem 7

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] usepackage("mathptmx"); 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 $\angle EFD = \frac{180^\circ - \angle ADE}{2}=10^\circ.$ Finally, we get $\angle AFE = 180^\circ - \angle EFD = \boxed{\textbf{(D) }170}$ degrees.

~MRENTHUSIASM

Solution 2 (same as Solution 1 but by another user)

Since $\angle CDE=110^\circ$ and $\angle ADC=90^\circ,$ we know that $\angle FDE=360^\circ-110^\circ-90^\circ=160^\circ.$ From the given, $DE=DF,$ so that means $\triangle DEF$ is isosceles. So, $\angle EFD=\left(\frac{180-160}{2}\right)^\circ=10^\circ.$ Hence, $\angle AFE=180^\circ-10^\circ=170^\circ$ since it is supplementary to $\angle EFD.$ This gives us answer choice $\boxed{\textbf{(D)}}.$

Aops-g5-gethsemanea2 (talk) 18:56, 22 November 2021 (EST) This solution conflicted with the above solution due to two people editing at the same time.

See Also

2021 Fall AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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