AMC10 2008 B

AMC10 2008 B · Q24

AMC10 2008 B · Q24. It mainly tests Angle chasing, Triangles (properties).

Quadrilateral ABCD has AB = BC = CD, $\angle ABC = 70^\circ$, and $\angle BCD = 170^\circ$. What is the degree measure of $\angle BAD$?

四边形ABCD有AB = BC = CD，$\angle ABC = 70^\circ$，$\angle BCD = 170^\circ$。$\angle BAD$的度量是多少度？

(A) 75 75

(B) 80 80

(D) 90 90

(E) 95 95

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Let $M$ be on the same side of line $BC$ as $A$, such that $\triangle BMC$ is equilateral. Then $\triangle ABM$ and $\triangle MCD$ are isosceles with $\angle ABM=10^\circ$ and $\angle MCD=110^\circ$. Hence $\angle AMB=85^\circ$ and $\angle CMD=35^\circ$. Therefore $\angle AMD=360^\circ-\angle AMB-\angle BMC-\angle CMD$ $=360^\circ-85^\circ-60^\circ-35^\circ=180^\circ.$ It follows that $M$ lies on $AD$ and $\angle BAD=\angle BAM=85^\circ$.

解答（C）：取点 $M$ 与 $A$ 在直线 $BC$ 的同侧，使得 $\triangle BMC$ 为正三角形。则 $\triangle ABM$ 与 $\triangle MCD$ 为等腰三角形，且 $\angle ABM=10^\circ$、$\angle MCD=110^\circ$。因此 $\angle AMB=85^\circ$、$\angle CMD=35^\circ$。所以 $\angle AMD=360^\circ-\angle AMB-\angle BMC-\angle CMD$ $=360^\circ-85^\circ-60^\circ-35^\circ=180^\circ.$ 由此可知 $M$ 在 $AD$ 上，且 $\angle BAD=\angle BAM=85^\circ$。

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