Problem: In triangle ABC,BDA B C, B DABC,BD is a median. CFC FCF intersects BDB DBD at EEE so that BE=EDB E=E DBE=ED. Point FFF is on ABABAB. Then, if BF=5,BABF=5, BABF=5,BA equals:
Answer Choices:
A. 101010
B. 121212
C. 151515
D. 202020
E. none of these Solution:
Let GGG be a point on EC so that FE=EGF E=E GFE=EG. Connect DDD with GGG. Then FDGB is a parallelogram.
therefore DG=5D G=5DG=5 AF=10A F=10AF=10 AB=15A B=15AB=15