AMC12 2009 A

AMC12 2009 A · Q20

AMC12 2009 A · Q20. It mainly tests Triangles (properties), Ratios in geometry.

Convex quadrilateral $ABCD$ has $AB = 9$ and $CD = 12$. Diagonals $AC$ and $BD$ intersect at $E$, $AC = 14$, and $\triangle AED$ and $\triangle BEC$ have equal areas. What is $AE$?

凸四边形 $ABCD$ 满足 $AB=9$ 且 $CD=12$。对角线 $AC$ 与 $BD$ 交于 $E$，$AC=14$，且 $\triangle AED$ 与 $\triangle BEC$ 的面积相等。求 $AE$。

(A) $\frac{9}{2}$ $\frac{9}{2}$

(B) $\frac{50}{11}$ $\frac{50}{11}$

(D) $\frac{17}{3}$ $\frac{17}{3}$

(E) 6 6

Answer

Correct choice: (E)

正确答案：（E）

Solution

Let $[ABC]$ denote the area of triangle $ABC$. $[AED] = [BEC]$, so $[ABD] = [AED] + [AEB] = [BEC] + [AEB] = [ABC]$. Since triangles $ABD$ and $ABC$ share a base, they also have the same height and thus $\overline{AB}||\overline{CD}$ and $\triangle{AEB}\sim\triangle{CED}$ with a ratio of $3: 4$. $AE = \frac {3}{7}\times AC$, so $AE = \frac {3}{7}\times 14 = 6\ \boxed{\textbf{(E)}}$.

设 $[ABC]$ 表示三角形 $ABC$ 的面积。由 $[AED]=[BEC]$，得 $[ABD]=[AED]+[AEB]=[BEC]+[AEB]=[ABC]$。由于三角形 $ABD$ 与 $ABC$ 共享同一底边，它们的高也相同，因此 $\overline{AB}||\overline{CD}$，并且 $\triangle{AEB}\sim\triangle{CED}$，相似比为 $3:4$。于是 $AE=\frac{3}{7}\times AC$，所以 $AE=\frac{3}{7}\times14=6\ \boxed{\textbf{(E)}}$。

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