Problem:
Let ABCD be a parallelogram with β ABC= 120β,AB=16 and BC=10. Extend CD through D to E so that DE=4. If BΛEΛ intersects AD at F, then FD is closest to
Answer Choices:
A. 1
B. 2
C. 3
D. 4
E. 5
Solution:
Since ADβ₯BC,β³FDE is similar to β³BCE. Hence
BCFDβ=CEDEβ or FD=CEDEββ
BC=4+164ββ
10=2
Note. The answer is independent of β ABC.
