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2021 Fall AMC 10A Problem 7

Problem 7 of 25EasierGeometry

As shown in the figure below, point EE lies in the opposite half-plane determined by line CDCD from point AA so that ∠CDE=110∘.\angle CDE = 110^\circ. Point FF lies on AD‾\overline{AD} so that DE=DF,DE=DF, and ABCDABCD is a square. What is the degree measure of ∠AFE?\angle AFE?

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Solution

Since ∠ADC=90∘,\angle ADC = 90^{\circ}, we get that ∠FDE=360∘−90∘−110∘ \angle FDE = 360^{\circ} - 90^{\circ} - 110^{\circ} =160∘.= 160^{\circ}. Also since △FDE\triangle FDE is isosceles, we get that ∠EFD=180∘−160∘2=10∘. \angle EFD = \dfrac{180^{\circ} - 160^{\circ}}{2} = 10^{\circ}. Finally, we get that ∠AFE=180∘−10∘=170∘. \angle AFE = 180^{\circ} - 10^{\circ} = 170^{\circ}. Thus, D is the correct answer.
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Tagged: angle chasing · isosceles triangle · square (geometry)

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