AMC12 2002 A

AMC12 2002 A · Q23

AMC12 2002 A · Q23. It mainly tests Angle chasing, Triangles (properties).

In triangle $ABC$, side $AC$ and the perpendicular bisector of $BC$ meet in point $D$, and $BD$ bisects $\angle ABC$. If $AD=9$ and $DC=7$, what is the area of triangle $ABD$?

在三角形 $ABC$ 中，边 $AC$ 与 $BC$ 的垂直平分线相交于点 $D$，且 $BD$ 平分 $\angle ABC$。若 $AD=9$ 且 $DC=7$，求三角形 $ABD$ 的面积。

(A) 14 14

(B) 21 21

(D) 14√5 14√5

(E) 28√5 28√5

Answer

Correct choice: (D)

正确答案：（D）

Solution

Looking at the triangle $BCD$, we see that its perpendicular bisector reaches the vertex, therefore implying it is isosceles. Let $x = \angle C$, so that $B=2x$ from given and the previous deducted. Then $\angle ABD=x, \angle ADB=2x$ because any exterior angle of a triangle has a measure that is the sum of the two interior angles that are not adjacent to the exterior angle. That means $\triangle ABD$ and $\triangle ACB$ are similar, so $\frac {16}{AB}=\frac {AB}{9} \Longrightarrow AB=12$. Then by using Heron's Formula on $ABD$ (with sides $12,7,9$), we have $[\triangle ABD]= \sqrt{14(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}$.

观察三角形 $BCD$，其垂直平分线经过顶点，因此该三角形为等腰三角形。设 $x = \angle C$，由已知以及前述结论可得 $B=2x$。于是 $\angle ABD=x, \angle ADB=2x$，因为三角形的任一外角等于与其不相邻的两个内角之和。这意味着 $\triangle ABD$ 与 $\triangle ACB$ 相似，所以 $\frac {16}{AB}=\frac {AB}{9} \Longrightarrow AB=12$。然后对 $ABD$（边长为 $12,7,9$）使用海伦公式，有 $[\triangle ABD]= \sqrt{14(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}$。

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