AMC12 2009 B

AMC12 2009 B · Q16

AMC12 2009 B · Q16. It mainly tests Angle chasing, Similarity.

Trapezoid $ABCD$ has $AD||BC$, $BD = 1$, $\angle DBA = 23^{\circ}$, and $\angle BDC = 46^{\circ}$. The ratio $BC: AD$ is $9: 5$. What is $CD$?

梯形 $ABCD$ 满足 $AD||BC$，$BD = 1$，$\angle DBA = 23^{\circ}$，且 $\angle BDC = 46^{\circ}$。比值 $BC: AD$ 为 $9: 5$。求 $CD$。

(A) $\frac{7}{9}$ $\frac{7}{9}$

(B) $\frac{4}{5}$ $\frac{4}{5}$

(D) $\frac{8}{9}$ $\frac{8}{9}$

(E) $\frac{14}{15}$ $\frac{14}{15}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Extend $\overline {AB}$ and $\overline {DC}$ to meet at $E$. Then \begin{align*} \angle BED &= 180^{\circ} - \angle EDB - \angle DBE\\ &= 180^{\circ} - 134^{\circ} -23^{\circ} = 23^{\circ}. \end{align*} Thus $\triangle BDE$ is isosceles with $DE = BD$. Because $\overline {AD} \parallel \overline {BC}$, it follows that the triangles $BCE$ and $ADE$ are similar. Therefore \[\frac 95 = \frac {BC}{AD} = \frac {CD + DE}{DE} = \frac {CD}{BD} + 1 = CD + 1,\] so $CD = \boxed{\frac 45}.$

延长 $\overline {AB}$ 和 $\overline {DC}$ 相交于 $E$。则 \begin{align*} \angle BED &= 180^{\circ} - \angle EDB - \angle DBE\\ &= 180^{\circ} - 134^{\circ} -23^{\circ} = 23^{\circ}. \end{align*} 因此 $\triangle BDE$ 为等腰三角形，且 $DE = BD$。由于 $\overline {AD} \parallel \overline {BC}$，可得三角形 $BCE$ 与 $ADE$ 相似。因此 \[\frac 95 = \frac {BC}{AD} = \frac {CD + DE}{DE} = \frac {CD}{BD} + 1 = CD + 1,\] 所以 $CD = \boxed{\frac 45}.$

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