$AD = 4,\; AB = 3,\; CD = 9\;$ What is the area of $\triangle AEC? $
I cant find a way out... Any hint will be helpful...
PS: This is a Problem from BDMO Regional.
2026-04-06 16:55:39.1775494539
$AD = 4\; AB = 3\; CD = 9\;$ What is the area of $\triangle AEC? $
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$\triangle ECD$ is similar to $\triangle EBA$, so $$\frac{AE}{DE} = \frac{AB}{DC} = \frac{3}{9}=\frac{1}{3}.$$ Since $AE+DE=AD=4$, we have $AE=1$ and $DE=3$. From here, subtracting the area of $\triangle CED$ from $\triangle ACD$ will give you the answer.