Difference between revisions of "1983 AHSME Problems/Problem 28"
(Created page with "Clearly since <math>[DBEF] = [ABE]</math> it follows that <math>[ADF] = [AFE]</math>. This implies that <math>AC \parallel DE</math> and so <math>\frac{AD}{DB} = \frac{CE}{EB}...") |
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Revision as of 08:08, 4 August 2017
Clearly since ![$[DBEF] = [ABE]$](http://latex.artofproblemsolving.com/3/0/1/301828b8af77c7c2cb4248d6cdb60ad8097c49e9.png)
it follows that ![$[ADF] = [AFE]$](http://latex.artofproblemsolving.com/a/2/9/a2953a78792495f0772305cfe55a6f520a0fe147.png)
. This implies that 
and so 
. Thus ![$[AEB] = \frac{3}{2+3} \cdot 10 = \boxed{6}$](http://latex.artofproblemsolving.com/b/a/a/baac58f4d561840e91d6cc8726b55a25b1c82d6c.png)
