In the diagram below, we have $\overline{AB}\parallel\overline{DE}$ and $BD = 24$. Find $CD$.

Answer 1

Since $\overline{AB}\parallel \overline{DE}$, we have $\triangle BCD \sim \triangle ECD$ by AA Similarity. Since $BD=24$, $BC:CD=24:CD$ or equivalently $BC:CD=CD:24$. Thus $\triangle BCD \sim \triangle ECD \sim \triangle BDC$. In particular, $\angle CDB \cong \angle CBD$, so $\triangle BCD$ is isosceles with $CD=BD=\boxed{24}$.

[asy]
pair A,B,C,D,E,EE;

A=(0,12);
B=(34,12);
C=(34,0);
D=(0,0);
EE=(24,0);
draw(A--B--C--D--cycle,linewidth(0.7));
draw(D--EE,linewidth(0.7));
label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D$",D,SW);
label("$D$",EE,S);
label("24",(A+B)/2,N);
label("24",(D+EE)/2,S);
[/asy]