# 1983 AHSME Problems/Problem 28

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Clearly since $[DBEF] = [ABE]$ it follows that $[ADF] = [AFE]$. This implies that $AC \parallel DE$ and so $\frac{AD}{DB} = \frac{CE}{EB} = \frac{2}{3}$. Thus $[AEB] = \frac{3}{2+3} \cdot 10 = \boxed{6}$