2017 AMC 10B Problem 15

The area of $\triangle CDA$ is $\dfrac{3\cdot4}{2}=6$. Since $E$ lies on $AC$, triangles $EAD$ and $CDA$ share the same altitude from $D$, so $[EAD]=[CDA]\cdot \dfrac{AE}{AC}$. By the Pythagorean Theorem, $AC=5$. Also $\triangle ABE\sim \triangle ACB$, so $\dfrac{AE}{AB}=\dfrac{AB}{AC}=\dfrac35$, giving $AE=\dfrac95$. Thus $\dfrac{AE}{AC}=\dfrac{9}{25}$. Therefore $[EAD]=6\cdot \dfrac{9}{25}=\dfrac{54}{25}$. Thus, E is the correct answer.