1995 AHSME Problems/Problem 8

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Problem

In $\triangle ABC$ , $\angle C = 90^\circ, AC = 6$ and $BC = 8$ . Points $D$ and $E$ are on $\overline{AB}$ and $\overline{BC}$ , respectively, and $\angle BED = 90^\circ$ . If $DE = 4$ , then $BD =$

$\mathrm{(A) \ 5 } \qquad \mathrm{(B) \ \frac {16}{3} } \qquad \mathrm{(C) \ \frac {20}{3} } \qquad \mathrm{(D) \ \frac {15}{2} } \qquad \mathrm{(E) \ 8 }$

Solution

[Asy_image]

$\triangle BDE$ is similar to $\triangle BAC$ , and thus $BD=10\cdot \dfrac{4}{6}=\dfrac{20}{3}\Rightarrow \boxed{\mathrm{(C)}}$ .

1995 AHSME (Problems)
Preceded by Problem 7	Followed by Problem 9
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1995 AHSME Problems/Problem 8

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Problem

Solution

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