AMC10 2011 B

AMC10 2011 B · Q17

AMC10 2011 B · Q17. It mainly tests Angle chasing, Triangles (properties).

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

在给定的圆中，直径 EB 平行于 DC，且 AB 平行于 ED。角 AEB 和 ABE 的比值为 4 : 5。角 BCD 的度数是多少？

(A) 120 120

(B) 125 125

(D) 135 135

(E) 140 140

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Angle $EAB$ is $90^\circ$ because it subtends a diameter. Therefore angles $BEA$ and $ABE$ are $40^\circ$ and $50^\circ$, respectively. Angle $DEB$ is $50^\circ$ because $AB$ is parallel to $ED$. Also, $\angle DEB$ is supplementary to $\angle CDE$, so $\angle CDE = 130^\circ$. Because $EB$ and $DC$ are parallel chords, $ED = BC$ and $EBCD$ is an isosceles trapezoid. Thus $\angle BCD = \angle CDE = 130^\circ$. OR Let $O$ be the center of the circle. Establish, as in the first solution, that $\angle EAB = 90^\circ$, $\angle BEA = 40^\circ$, $\angle ABE = 50^\circ$, and $\angle DEB = 50^\circ$. Thus $AD$ is a diameter and $\angle AOE = 100^\circ$. By the Inscribed Angle Theorem $$ \angle BCD=\frac12(\angle BOA+\angle AOE+\angle EOD)=\frac12(80^\circ+100^\circ+80^\circ)=130^\circ. $$

答案（C）：由于弦 $EA$ 所对的是直径，所以 $\angle EAB$ 为 $90^\circ$。因此，$\angle BEA$ 和 $\angle ABE$ 分别为 $40^\circ$ 和 $50^\circ$。因为 $AB \parallel ED$，所以 $\angle DEB = 50^\circ$。另外，$\angle DEB$ 与 $\angle CDE$ 互为补角，因此 $\angle CDE = 130^\circ$。由于 $EB$ 与 $DC$ 是平行弦，所以 $ED = BC$，从而 $EBCD$ 是等腰梯形。于是 $\angle BCD = \angle CDE = 130^\circ$。或者设 $O$ 为圆心。与第一种解法同理可得：$\angle EAB = 90^\circ$，$\angle BEA = 40^\circ$，$\angle ABE = 50^\circ$，以及 $\angle DEB = 50^\circ$。因此 $AD$ 是直径，且 $\angle AOE = 100^\circ$。由圆周角定理， $$ \angle BCD=\frac12(\angle BOA+\angle AOE+\angle EOD)=\frac12(80^\circ+100^\circ+80^\circ)=130^\circ. $$

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