AMC10 2018 A

AMC10 2018 A · Q24

AMC10 2018 A · Q24. It mainly tests Angle chasing, Triangles (properties).

Triangle ABC with AB = 50 and AC = 10 has area 120. Let D be the midpoint of AB, and let E be the midpoint of AC. The angle bisector of ∠BAC intersects DE and BC at F and G, respectively. What is the area of quadrilateral FDBG ?

三角形 ABC 有 AB = 50 和 AC = 10，面积为 120。设 D 为 AB 的中点，E 为 AC 的中点。∠BAC 的角平分线交 DE 和 BC 于 F 和 G 分别。何处四边形 FDBG 的面积？

(A) 60 60

(B) 65 65

(D) 75 75

(E) 80 80

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Because $AB$ is $\frac{5}{6}$ of $AB+AC$, it follows from the Angle Bisector Theorem that $DF$ is $\frac{5}{6}$ of $DE$, and $BG$ is $\frac{5}{6}$ of $BC$. Because trapezoids $FDBG$ and $EDBC$ have the same height, the area of $FDBG$ is $\frac{5}{6}$ of the area of $EDBC$. Furthermore, the area of $\triangle ADE$ is $\frac{1}{4}$ of the area of $\triangle ABC$, so its area is $30$, and the area of trapezoid $EDBC$ is $120-30=90$. Therefore the area of quadrilateral $FDBG$ is $\frac{5}{6}\cdot 90=75$. Note: The figure (not drawn to scale) shows the situation in which $\angle ACB$ is acute. In this case $BC\approx 59.0$ and $\angle BAC\approx 151^\circ$. It is also possible for $\angle ACB$ to be obtuse, with $BC\approx 41.5$ and $\angle BAC\approx 29^\circ$. These values can be calculated using the Law of Cosines and the sine formula for area.

答案（D）：因为$AB$是$AB+AC$的$\frac{5}{6}$，由角平分线定理可知$DF$是$DE$的$\frac{5}{6}$，且$BG$是$BC$的$\frac{5}{6}$。由于梯形$FDBG$与$EDBC$具有相同的高，$FDBG$的面积是$EDBC$面积的$\frac{5}{6}$。另外，$\triangle ADE$的面积是$\triangle ABC$面积的$\frac{1}{4}$，所以其面积为$30$，而梯形$EDBC$的面积为$120-30=90$。因此四边形$FDBG$的面积为$\frac{5}{6}\cdot 90=75$。注：图（未按比例绘制）展示了$\angle ACB$为锐角的情形。在这种情况下$BC\approx 59.0$且$\angle BAC\approx 151^\circ$。$\angle ACB$也可能为钝角，此时$BC\approx 41.5$且$\angle BAC\approx 29^\circ$。这些数值可用余弦定理以及面积的正弦公式计算。

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