# 1990 AHSME Problems/Problem 4

## Problem $[asy] draw((0,0)--(16,0)--(21,5*sqrt(3))--(5,5*sqrt(3))--cycle,dot); draw((5,5*sqrt(3))--(1,5*sqrt(3))--(16,0),dot); MP("A",(0,0),S);MP("B",(16,0),S);MP("C",(21,5sqrt(3)),NE);MP("D",(5,5sqrt(3)),N);MP("E",(1,5sqrt(3)),N); MP("16",(8,0),S);MP("10",(18.5,5sqrt(3)/2),E);MP("4",(3,5sqrt(3)),N); dot((4,4sqrt(3))); MP("F",(4,4sqrt(3)),dir(210)); [/asy]$

Let $ABCD$ be a parallelogram with $\angle{ABC}=120^\circ, AB=16$ and $BC=10.$ Extend $\overline{CD}$ through $D$ to $E$ so that $DE=4.$ If $\overline{BE}$ intersects $\overline{AD}$ at $F$, then $FD$ is closest to $\text{(A) } 1\quad \text{(B) } 2\quad \text{(C) } 3\quad \text{(D) } 4\quad \text{(E) } 5$

## Solution $DFE$ and $AFB$ are similar triangles, so $FD$ is one quarter the length of the corresponding side $AF$. Thus it is one fifth of the length of $AD$, which means $\fbox{B}$

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