AMC10 2009 B

AMC10 2009 B · Q9

AMC10 2009 B · Q9. It mainly tests Angle chasing, Triangles (properties).

Segment $BD$ and $AE$ intersect at $C$, as shown, $AB = BC = CD = CE$, and $\angle A = \frac{3}{2}\angle B$. What is the degree measure of $\angle D$?

线段 $BD$ 和 $AE$ 相交于 $C$，如图所示，$AB = BC = CD = CE$，且 $\angle A = \frac{3}{2}\angle B$。$\angle D$ 的度数是多少？

(A) 52.5 52.5

(B) 55 55

(D) 60 60

(E) 62.5 62.5

Answer

Correct choice: (A)

正确答案：（A）

Solution

Answer (A): Because $\triangle ABC$ is isosceles, $\angle A=\angle C$. Because $\angle A=\frac{5}{2}\angle B$, we have $\frac{5}{2}\angle B+\frac{5}{2}\angle B+\angle B=180^\circ$, so $\angle B=30^\circ$. Therefore $\angle ACB=\angle DCE=75^\circ$. Because $\triangle CDE$ is isosceles, $2\angle D+75^\circ=180^\circ$, so $\angle D=52.5^\circ$.

答案（A）：因为 $\triangle ABC$ 是等腰三角形，所以 $\angle A=\angle C$。又因为 $\angle A=\frac{5}{2}\angle B$，所以 $\frac{5}{2}\angle B+\frac{5}{2}\angle B+\angle B=180^\circ$，从而 $\angle B=30^\circ$。因此 $\angle ACB=\angle DCE=75^\circ$。因为 $\triangle CDE$ 是等腰三角形，所以 $2\angle D+75^\circ=180^\circ$，从而 $\angle D=52.5^\circ$。

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