Let ABCD be a parallelogram with ∠ ABC=120°, AB=16, and BC=10. Extend CD through D to E so that DE=4. If BE intersects AD at F, then FD is closest to

1990 AMC 12 Problem 4

Attempt Problem 4 of the 1990 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1990 AMC 12 solutions, or check the answer key.

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Let

ABCD

be a parallelogram with

\angle ABC=120^\circ,

AB=16,

and

BC=10.

Extend

\overline{CD}

through

D

E

so that

DE=4.

\overline{BE}

intersects

\overline{AD}

F,

then

FD

is closest to

1

2

3

4

5

Use that

\overline{AB}\parallel\overline{ED}

Triangles

ABF

and

DEF

are similar

Because

AB\parallel ED,

triangles

ABF

and

DEF

are similar. Hence

\frac{AF}{FD}=\frac{AB}{DE}=\frac{16}{4}=4.

Since

AF+FD=AD=BC=10,

we get

5FD=10,

FD=2.

Thus the correct answer is B.