AMC8 1996

AMC8 1996 · Q24

AMC8 1996 · Q24. It mainly tests Angle chasing, Triangles (properties).

The measure of angle $ABC$ is $50^\circ$, $\overline{AD}$ bisects angle $BAC$, and $\overline{DC}$ bisects angle $BCA$. The measure of angle $ADC$ is

角 $ABC$ 的度数为 $50^\circ$，$\overline{AD}$ 平分角 $BAC$，$\overline{DC}$ 平分角 $BCA$。角 $ADC$ 的度数是

(A) $90^\circ$ $90^\circ$

(B) $100^\circ$ $100^\circ$

(D) $122.5^\circ$ $122.5^\circ$

(E) $125^\circ$ $125^\circ$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Answer (C): Since the sum of the measures of the angles of a triangle is 180°, in triangle $ABC$ it follows that $\angle BAC+\angle BCA=180^\circ-50^\circ=130^\circ.$ The measures of angles $DAC$ and $DCA$ are half that of angles $BAC$ and $BCA$, respectively, so $\angle DAC+\angle DCA=\frac{130^\circ}{2}=65^\circ.$ In triangle $ACD$, we have $\angle ADC=180^\circ-65^\circ=115^\circ.$

答案（C）：由于三角形内角和为 $180^\circ$，在三角形 $ABC$ 中有 $\angle BAC+\angle BCA=180^\circ-50^\circ=130^\circ.$ 角 $DAC$ 和 $DCA$ 的度数分别是角 $BAC$ 和 $BCA$ 的一半，因此 $\angle DAC+\angle DCA=\frac{130^\circ}{2}=65^\circ.$ 在三角形 $ACD$ 中，有 $\angle ADC=180^\circ-65^\circ=115^\circ.$

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