AMC8 1995

AMC8 1995 · Q13

AMC8 1995 · Q13. It mainly tests Angle chasing, Triangles (properties).

In the figure, ∠A, ∠B and ∠C are right angles. If ∠AEB = 40° and ∠BED = ∠BDE, then ∠CDE =

在图中，∠A、∠B和∠C是直角。如果∠AEB = 40°且∠BED = ∠BDE，则∠CDE =

(A) 75° 75°

(B) 80° 80°

(D) 90° 90°

(E) 95° 95°

Answer

Correct choice: (E)

正确答案：（E）

Solution

Answer (E): In $\triangle BDE$, $\angle BED + \angle BDE + \angle B = 180^\circ$. Since $\angle BED = \angle BDE$ and $\angle B = 90^\circ$, it follows that $\angle BED = \angle BDE = 45^\circ$. In $\triangle AEF$, $\angle A + \angle AEF + \angle AFE = 180^\circ$. Since $\angle A = 90^\circ$ and $\angle AEF = 40^\circ$, it follows that $\angle AFE = 50^\circ$. Consequently $\angle BFG = 50^\circ$ in $\triangle BFG$ and, since $\angle B = 90^\circ$, it follows that $\angle BGF = 40^\circ$. Consequently $\angle CGD = 40^\circ$ in $\triangle CDG$, and since $\angle C = 90^\circ$, it follows that $\angle CDG = 50^\circ$. Thus $\angle CDE = 50^\circ + 45^\circ = 95^\circ$.

答案（E）：在$\triangle BDE$中，$\angle BED+\angle BDE+\angle B=180^\circ$。由于$\angle BED=\angle BDE$且$\angle B=90^\circ$，可得$\angle BED=\angle BDE=45^\circ$。在$\triangle AEF$中，$\angle A+\angle AEF+\angle AFE=180^\circ$。由于$\angle A=90^\circ$且$\angle AEF=40^\circ$，可得$\angle AFE=50^\circ$。因此在$\triangle BFG$中$\angle BFG=50^\circ$，又因为$\angle B=90^\circ$，可得$\angle BGF=40^\circ$。因此在$\triangle CDG$中$\angle CGD=40^\circ$，又因为$\angle C=90^\circ$，可得$\angle CDG=50^\circ$。所以$\angle CDE=50^\circ+45^\circ=95^\circ$。

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