AMC8 2000

AMC8 2000 · Q24

AMC8 2000 · Q24. It mainly tests Angle chasing, Triangles (properties).

If $\angle A = 20^\circ$ and $\angle AFG = \angle AGF$, then $\angle B + \angle D =$

若$\angle A = 20^\circ$且$\angle AFG = \angle AGF$，则$\angle B + \angle D =$

(A) 48° 48°

(B) 60° 60°

(D) 80° 80°

(E) 90° 90°

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Since $\angle AFG = \angle AGF$ and $\angle GAF + \angle AFG + \angle AGF = 180^\circ$, we have $20^\circ + 2(\angle AFG) = 180^\circ$. So $\angle AFG = 80^\circ$. Also, $\angle AFG + \angle BFD = 190^\circ$, so $\angle BFD = 100^\circ$. The sum of the angles of $\triangle BFD$ is $180^\circ$, so $\angle B + \angle D = 80^\circ$. Note: In $\triangle AFG$, $\angle AFG = \angle B + \angle D$. In general, an exterior angle of a triangle equals the sum of its remote interior angles. For example, in $\triangle GAF$, $\angle x = \angle GAF + \angle AGF$. Note that, as in Problem 13, some texts use different symbols to represent an angle and its degree measure.

答案（D）：由于 $\angle AFG = \angle AGF$ 且 $\angle GAF + \angle AFG + \angle AGF = 180^\circ$，我们有 $20^\circ + 2(\angle AFG) = 180^\circ$。所以 $\angle AFG = 80^\circ$。另外，$\angle AFG + \angle BFD = 190^\circ$，因此 $\angle BFD = 100^\circ$。$\triangle BFD$ 的内角和为 $180^\circ$，所以 $\angle B + \angle D = 80^\circ$。注：在 $\triangle AFG$ 中，$\angle AFG = \angle B + \angle D$。一般地，三角形的一个外角等于与之不相邻的两个内角之和。例如，在 $\triangle GAF$ 中，$\angle x = \angle GAF + \angle AGF$。还要注意：如同第 13 题所示，有些教材会用不同的符号来表示“角”及其“度数”。

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