2011 AMC 10B Problem 17

Work from the top triangle down to $C$, one angle at a time.

1. Triangle $AEB$. $EB$ is a diameter, and any angle drawn from a diameter to a point on the circle is a right angle. So $\angle EAB = 90^\circ$.
2. The other two angles of triangle $AEB$ add to $180 - 90 = 90$ degrees, in the ratio $4 : 5$. Split $90$ into $9$ equal parts of $10$: $\angle AEB = 40^\circ$ and $\angle ABE = 50^\circ$.
3. Move the $50^\circ$ down. $AB$ is parallel to $ED$, and $EB$ crosses both. The angles $\angle ABE$ (above $EB$) and $\angle BED$ (below $EB$) are alternate angles, so they are equal: $\angle BED = 50^\circ$.
4. The bottom shape is symmetric. $EDCB$ has $DC$ parallel to $EB$, and all four corners are on the circle. A shape like this is the same on its left and right halves (fold it along the line through the center perpendicular to $EB$). So $\angle EBC = \angle BED = 50^\circ$.
5. Finish with the parallel lines. $DC$ is parallel to $EB$, and $BC$ crosses both. The two angles on the same side, $\angle BCD$ and $\angle CBE$, add to $180^\circ$. So
$$\angle BCD = 180^\circ - 50^\circ = 130^\circ.$$

The answer is $\boxed{\textbf{(C)}\ 130}$.

Three facts are chained together: an angle standing on a diameter is $90^\circ$; parallel lines let you copy an angle to a new spot; and a four-sided shape inside a circle with two parallel sides is symmetric. In circle problems, look for a diameter first, because it hands you a right angle for free.

Once you know $\angle BED = 50^\circ$, use the rule for a four-sided shape with all corners on a circle: opposite angles add to $180^\circ$. In $EDCB$, the angles at $E$ and $C$ are opposite, so $\angle BCD = 180^\circ - 50^\circ = 130^\circ$.