Problem: In △ABC,E is the midpoint of side BC and D is on side AC. If the length of AC is 1 and ∠BAC=60∘,∠ABC=100∘,∠ACB=20∘ and ∠DEC=80∘, then the area of △ABC plus twice the area of △CDE equals
Answer Choices:
A. 41cos10∘
B. 83
C. 41cos40∘
D. 41cos50∘
E. 81
Solution:
Let F be the point on the extension of side AB past B for which AF=1. Since AF=AC and ∠FAC=60∘,△ACF is equilateral. Let G be the point on line segment BF for which ∠BCG=20∘. Then △BCG is similar to △DCE and BC=2(EC). Also △FGC is congruent to △ABC. Therefore,
Area △ACF=( Area △ABC+ Area △GCF)+ Area △BCG
43=2 Area △ABC+4 Area △CDE83= Area △ABC+2 Area △CDE
