In the given circle, the diameter $\overline{EB}$ is parallel to $\overline{DC}$, and $\overline{AB}$ is parallel to $\overline{ED}$. The angles $AEB$ and $ABE$ are in the ratio $4 : 5$. What is the degree measure of angle $BCD$?


$\textbf{(A)}\ 120 \qquad\textbf{(B)}\ 125 \qquad\textbf{(C)}\ 130 \qquad\textbf{(D)}\ 135 \qquad\textbf{(E)}\ 140$
We can let $\angle AEB$ be $4x$ and $\angle ABE$ be $5x$ because they are in the ratio $4 : 5$. When an [inscribed angle](https://artofproblemsolving.com/wiki/index.php/Inscribed_angle) contains the [diameter](https://artofproblemsolving.com/wiki/index.php/Diameter), the inscribed angle is a [right angle](https://artofproblemsolving.com/wiki/index.php/Right_angle). Therefore by triangle sum theorem, $4x+5x+90=180 \longrightarrow x=10$ and $\angle ABE = 50$.
$\angle ABE = \angle BED$ because they are alternate interior angles and $\overline{AB} \parallel \overline{ED}$. Opposite angles in a [cyclic](https://artofproblemsolving.com/wiki/index.php/Cyclic) quadrilateral are [supplementary](https://artofproblemsolving.com/wiki/index.php/Supplementary), so $\angle BED + \angle BCD = 180$. Use substitution to get $\angle ABE + \angle BCD = 180 \longrightarrow 50 + \angle BCD = 180 \longrightarrow \angle BCD = \boxed{\textbf{(C)} 130}$