1995 AHSME Problems/Problem 8

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

$[asy] size(120); defaultpen(0.7); pair A = (0,6), B = (8,0), C= (0,0), D = (56/15,3.2), E = (4/3,0), F = (0, 3), G = (38/15,1.6); draw(A--B--E--D--E--B--C--A--B--cycle); label("$A$",A,W); label("$B$",B,E); label("$C$",C,SW); label("$D$",D,NE); label("$E$",E,S); label("$6$",F,W); label("$4$",G,NW); [/asy]$

Since $\triangle BED$ is similar to $\triangle ABC$ , it is a 3-4-5 triangle, and thus $BD=4\cdot \dfrac{4}{3}=\dfrac{16}{3}\Rightarrow \boxed{\mathrm{(B)}}$ .

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1995 AHSME Problems/Problem 8

Problem

Solution

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