2011 AMC 10B Problem 17

Since $EB$ is a diameter, $\angle EAB = 90^\circ$. The other two angles of triangle $AEB$ sum to $90^\circ$ in ratio $4 : 5$, so $\angle AEB = 40^\circ$ and $\angle ABE = 50^\circ$.

Because $AB \parallel ED$ with transversal $EB$, alternate interior angles give $\angle BED = \angle ABE = 50^\circ$.

Points $E, D, C, B$ lie on the circle in that order (with $DC \parallel EB$, the chord $DC$ sits on one side of the diameter and $C$ is between $D$ and $B$ along the arc). So $EDCB$ is a cyclic quadrilateral, and its opposite angles at $E$ and $C$ are supplementary:
$$\angle BCD = 180^\circ - \angle DEB = 180^\circ - 50^\circ = 130^\circ.$$

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

Three standard facts chain together: Thales' theorem converts the diameter into a right angle, parallel chords transfer angles along a transversal, and a cyclic quadrilateral links an angle to its opposite. Equivalently, $\angle BED$ intercepts arc $BD$ ($100^\circ$) while $\angle BCD$ intercepts the rest of the circle ($260^\circ$).

$EDCB$ is an inscribed trapezoid ($DC \parallel EB$), hence isosceles, so $\angle EBC = \angle BED = 50^\circ$. Co-interior angles along leg $BC$ then give $\angle BCD = 180^\circ - 50^\circ = 130^\circ$.

• Concluding $\angle BCD = \angle BED = 50^\circ$ by "same arc", when $C$ lies on the arc between $B$ and $D$, so the two angles are supplementary, not equal.
• Swapping the ratio and taking $\angle ABE = 40^\circ$, which leads to $140^\circ$.