1976 AHSME Problems/Problem 9

Solution

Let $[ABC]$ denote the area of $ABC$ Since $F$ is the midpoint of $|BC|$ , we have $[ABF] = [ACF] = \frac{96}{2} = 48$ $D$ is the midpoint of $|AB|$ , so $[ADF] = [BDF] = \frac{48}{2} = 24$ Now we just need to find $[DEF]$ , since $[AEF] = [ADF] + [DEF]$ In $\triangle DBF$ , $E$ is the midpoint of $DB$ , so $[DEF] = [EBF] = \frac{24}{2} = 12$ Thus, $[AEF] = [ADF] + [DEF] = 24+12 = 36$ $\boxed{D}$

~JustinLee2017

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1976 AHSME Problems/Problem 9

Solution