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!

 A: Suppose a triangle $T_1$ has two sides $m,n$, and that the angle between those two sides is $\theta$. If we create a second triangle $T_2$ with two sides $\alpha \cdot m, \beta \cdot n$ and keep the angle between them as $\theta$, then
Area $(T_2) \;$ = $\; \alpha \beta \;\cdot \;$ Area $(T_1)$
This can easily be seen, by first keeping one of the sides (say $n$) fixed, and multiplying the other (in this case, $\alpha \cdot m$). This intermediate triangle would have the same height as the original, but the base would get multiplied by $\alpha$, so the area would also be multiplied by $\alpha$. Then the above result follows by multiplying the other side by its factor.
In your question, side AB gets multiplied by $\frac{100}{160}=\frac{5}{8}$, whereas the area gets multiplied by $\frac{3}{5}$. Thus the second multiplier (call it $x$) must satisfy
$$\frac{5}{8} x = \frac{3}{5}$$
which yields $x=\frac{24}{25}$.
Thus the length of AE will be
$$\frac{24}{25} (190) = \boxed{182.4}$$
A: As pointed out by Jaap in the comments, $\triangle AED$ may not be isosceles so $h_2$ may not be $\sqrt{|ED|^2-(\color{red}{100/2})^2}$.
Label the base of $h_1$ as $N$. Let $EM$ be the perpendicular on $AD$ from $E$. Then $|EM|=h_2$. Then$$\text{ar}\triangle ADE=\frac12\times h_2\times |AD|=\frac35(13786.95)$$giving $h_2=165.44$. Note that $\triangle AEM\sim\triangle ACN$, thus$$\frac{h_2}{h_1}=\frac{|AE|}{|AC|}$$giving $|AE|=182.4$.
A: 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 $
