AMC10 2007 A

AMC10 2007 A · Q8

AMC10 2007 A · Q8. It mainly tests Angle chasing, Triangles (properties).

Triangles $ABC$ and $ADC$ are isosceles with $AB = BC$ and $AD = DC$. Point $D$ is inside $\triangle ABC$, $\angle ABC = 40^\circ$, and $\angle ADC = 140^\circ$. What is the degree measure of $\angle BAD$?

等腰三角形$ABC$和$ADC$满足$AB = BC$和$AD = DC$。点$D$在$\triangle ABC$内部，$\angle ABC = 40^\circ$，$\angle ADC = 140^\circ$。$\angle BAD$的度数是多少？

(A) 20 20

(B) 30 30

(D) 50 50

(E) 60 60

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Because $\triangle ABC$ is isosceles, $\angle BAC=\frac{1}{2}(180^\circ-\angle ABC)=70^\circ$. Similarly, $\angle DAC=\frac{1}{2}(180^\circ-\angle ADC)=20^\circ$. Thus $\angle BAD=\angle BAC-\angle DAC=50^\circ$. OR Because $\triangle ABC$ and $\triangle ADC$ are isosceles triangles and $\overline{BD}$ bisects $\angle ABC$ and $\angle ADC$, applying the Exterior Angle Theorem to $\triangle ABD$ gives $\angle BAD=70^\circ-20^\circ=50^\circ$.

答案（D）：因为$\triangle ABC$是等腰三角形，$\angle BAC=\frac{1}{2}(180^\circ-\angle ABC)=70^\circ$。同理， $\angle DAC=\frac{1}{2}(180^\circ-\angle ADC)=20^\circ$。因此$\angle BAD=\angle BAC-\angle DAC=50^\circ$。或者因为$\triangle ABC$和$\triangle ADC$都是等腰三角形，且$\overline{BD}$平分$\angle ABC$和$\angle ADC$，对$\triangle ABD$应用外角定理可得$\angle BAD=70^\circ-20^\circ=50^\circ$。

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