Problem: Point F is taken on the extension of side AD of parallelogram ABCD. BF intersects diagonal AC at E and side DC at G. If EF=32 and GF=24, then BE equals:
Answer Choices:
A. 4
B. 8
C. 10
D. 12
E. 16
Solution:
Let BE=x,DG=y,AB=b. Since △BEA∼△GEC,
x8=bb−y,b−y=x8 b,y=b−x8 b=bbx(x−8)
Since △FDG∼△BCG,x+824=b−yy, x+824=x⋅x8bb(x−8)=8x−8,x2−64=192,x=16
Note: Try to prove generally that BE1=BG1+BF1
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