2011 AMC 10B Problem 17

Since $EB$ is a diameter, $\angle EAB=90^\circ$. The ratio $\angle AEB:\angle ABE=4:5$ then gives $\angle AEB=40^\circ$ and $\angle ABE=50^\circ$. Because $AB\parallel ED$, $\angle DEB=50^\circ$. Since $EB\parallel DC$, quadrilateral $EBCD$ is an isosceles trapezoid, so $\angle BCD=\angle CDE$. Angles $DEB$ and $CDE$ are supplementary, so $\angle CDE=180^\circ-50^\circ=130^\circ$. Hence $\angle BCD=130^\circ$. Thus, C is the correct answer.