Problem:
As shown in the figure below, point E lies in the opposite half-plane determined by line CD from point A so that ∠CDE=110∘. Point F lies on AD so that DE=DF, and ABCD is a square. What is the degree measure of ∠AFE ?
Answer Choices:
A. 160
B. 164
C. 166
D. 170
E. 174
Solution:
Note that ∠EDF=360∘−∠ADC−∠CDE=360∘−90∘−110∘=160∘. Because △DEF is isosceles, angles DEF and DFE have an equal measure of 2180∘−160∘=10∘. Hence ∠AFE= 180∘−∠DFE=180∘−10∘=(D)170∘.
The problems on this page are the property of the MAA's American Mathematics Competitions