Find a side of smaller triangle 
$\triangle ABC$ has side $\overline{AB}=160 cm$, $\overline{BC}=190 cm$, and $\overline{CA}=190 cm$.
Point $D$ is along side $\overline{AB}$ and $\overline{AD} = 100 cm$. Point $E$ is along side $\overline{CA}$.


Determine the length of $\overline{AE}$ if the area of $\triangle ADE$ is three-fifths the area of $\triangle ABC$.

Is my solution correct? Thanks in advance!


As indicated in comment, you're assuming $\triangle AED$ is isosceles (i.e., $DE$ || $BC$) which need not be true.
If you know formula for area of a $\triangle$ is also given as
$$ \dfrac{1}{2}AB\cdot AC\sin A$$
then using this one can directly solve
$$ \dfrac{1}{2}AD\cdot AE\sin A = \dfrac{3}{5} \cdot\dfrac{1}{2}AB\cdot AC\sin A$$
$$ \Rightarrow AE = \dfrac{3}{5}\dfrac{AB\cdot AC}{AD}$$
without any use of trigonometry.
Otherwise,
$$ \dfrac{1}{2}AD\cdot h_2 = \dfrac{3}{5} \cdot\dfrac{1}{2}AB\cdot h_1$$
$$ \Rightarrow h_2 = \dfrac{24}{25}h_1$$
and since the two right triangles formed by $h_1$, $h_2$ and containing $\angle A$ are similar
$$ \dfrac{AE}{AC} = \dfrac{h_2}{h_1}$$
$$ \Rightarrow AE = \dfrac{24}{25}AC$$
Either should give $AE = 182.4 $
