5. In trapezoid ABCD, AB∥CD,AB=3CD,E is the midpoint of diagonal AC, and line BE intersects AD at F. Then the value of AF:FD is:
Pick one
Solution
5. C.
Draw CG//BF intersecting the extension of AD at G (as shown in the figure). From AB//CD,CG//BF, we get △ABF∽△DCG, thus AF:DG=AB ~ DC=3:1. Also, by CG//BF, FD=εDC, hence AF:Fl=−32=23.
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