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 $x= \boxed{20}$. [asy]
unitsize(0.6inch);
pair A,B,C,D,E;
A = (0,0);
B = (4,0);
C= (4,2);
D = (8,2);
E = (0,-4);
draw(A--D--E--cycle,linewidth(1));
draw(D--C--B,dashed);
label("$A$",A,WSW);
label("$B$",B,SE);
label("$C$",C,NE);
label("$D$",D,ESE);
label("$E$",E,SWS);
label("$20$",(A+E)/2,S);
label("$20$",(C+D)/2,SE);
[/asy]
In the diagram below, we have $\overline{BC}\parallel\overline{DE}$ and $AE = 20$. Find $AC$.
1 answer