As shown in the figure, $ABCD$ is a parallelogram. $DE=3$, $DC=2$, find $BE$.
Below is what I have: $$AB=CD=2$$ $$AB^2=BE\times BC \implies 4=BE\times BC$$ $$AD\parallel BC \implies AE=CD=2$$ $$\angle ABE = \angle AEB = \angle ADC$$
That's all I found. How should I move on?
Draw AE and let the intersection of extended AB and DE be K. As $BK\parallel CD, \triangle DEC \sim \triangle EBK$. As $\frac{DC}{DE}=\frac{2}{3}$, You can label $EK=3x$ and $BK=2x$
As you found that $AE=CD$ I am not going to prove it again.
Also, you can see that $\triangle AEK \sim \triangle ADK$. By similarity, you can find the following relationship.
$\frac{2+2x}{3+3x}=\frac{3x}{2+2x}$
By simplifying
$(2+2x)^2=3x(3+3x)$
$x=\frac{4}{5}$ Wolfram Alpha
Again by the similarity of $\triangle AEK \sim \triangle DAK$ you get,
$\frac{2}{AD}=\frac{2+2x}{3+3x}$
Substituting the value for x you get,
$AD=3$ Wolfram Alpha
As $BE \parallel AD, \triangle EBK \sim \triangle ADK$ you get the following relationships,
$\frac{BE}{AD}=\frac{2x}{2+2x}$
By substituting values for $AD$ and $x$
$\boxed{BE=\frac{4}{3}}$ Wolfram Alpha