Difference between revisions of "1995 AHSME Problems/Problem 8"
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<math>\triangle BDE</math> is similar to <math>\triangle BAC</math> by angle-angle similarity since <math>E=C = 90^\circ</math> and <math>B=B</math>, and thus <math>\frac{BD}{BA} = \frac{DE}{AC}</math>. | <math>\triangle BDE</math> is similar to <math>\triangle BAC</math> by angle-angle similarity since <math>E=C = 90^\circ</math> and <math>B=B</math>, and thus <math>\frac{BD}{BA} = \frac{DE}{AC}</math>. | ||
| − | Solving the above for <math>BD</math>, we get <math>BD=\frac{BA\cdot DE | + | Solving the above for <math>BD</math>, we get <math>BD=\frac{BA\cdot DE}{AC} = 10\cdot \dfrac{4}{6}=\dfrac{20}{3}\Rightarrow \boxed{\mathrm{(C)}}</math>. |
==See also== | ==See also== | ||
Revision as of 00:18, 18 August 2011
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 a 
right triangle with hypotenuse 
.
is similar to by angle-angle similarity since 
and 
, and thus 
.
Solving the above for 
, we get 
.
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 | ||
