+1030 In the diagram below, we have $\overline{BC}\parallel\overline{DE}$ and $AE = 20$. Find $AC$.
+1484 Let $x = AC$. Since $\overline{BC}\parallel\overline{DE}$, we have that $\angle BAD$ and $\angle BCD$ are corresponding angles, and thus congruent. Therefore, $\triangle ABD \sim \triangle DCB$ by AA similarity.
From the similarity, we have \[
\frac{AB}{BD}=\frac{BD}{CB}\quad\Rightarrow\quad \frac{x}{20}=\frac{20}{x+20}.
\] Cross-multiplying gives $x^2+20x = 20^2$ so $x^2+20x-400=0$. Factoring the quadratic gives $(x-20)(x+40) = 0$, so $x= \boxed{20}$.
