Difference between revisions of "1995 AHSME Problems/Problem 8"
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| − | + | <math>\triangle BDE</math> is similar to <math>\triangle ABC</math>, and thus <math>BD=10\cdot \dfrac{4}{6}=\dfrac{20}{3}\Rightarrow \boxed{\mathrm{(C)}}</math>. | |
==See also== | ==See also== | ||
Revision as of 09:30, 9 August 2008
Problem
In 
, 
and 
. Points 
and 
are on 
and 
, respectively, and 
. If 
, then 
Solution
![[asy] size(120); defaultpen(0.7); pair A = (0,6), B = (8,0), C= (0,0), D = (8/3,4), E = (8/3,0), 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]](http://latex.artofproblemsolving.com/6/f/9/6f93ed41efdeef03b9969d66b7ef6567ccabc463.png)
is similar to , and thus 
.
See also
| 1995 AHSME (Problems • Answer Key • Resources) | ||
| 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 | ||
| All AHSME Problems and Solutions | ||
