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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 $BD =$

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

Solution

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

See also

1995 AHSME (Problems)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/1995_AHSME_Problems/Problem_8"

Category: Introductory Geometry Problems

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