AMC10 2009 A

AMC10 2009 A · Q23

AMC10 2009 A · Q23. It mainly tests Triangles (properties), Similarity.

Convex quadrilateral ABCD has AB = 9 and CD = 12. Diagonals AC and BD intersect at E, AC = 14, and ΔAED and ΔBEC have equal areas. What is AE?

凸四边形ABCD有$AB=9$且$CD=12$。对角线AC和BD交于E，$AC=14$，且$\triangle AED$与$\triangle BEC$面积相等。$AE$是多少？

(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

Answer (E): Because $\triangle AED$ and $\triangle BEC$ have equal areas, so do $\triangle ACD$ and $\triangle BCD$. Side $CD$ is common to $\triangle ACD$ and $\triangle BCD$, so the altitudes from $A$ and $B$ to $CD$ have the same length. Thus $AB \parallel CD$, so $\triangle ABE$ is similar to $\triangle CDE$ with similarity ratio \[ \frac{AE}{EC}=\frac{AB}{CD}=\frac{9}{12}=\frac{3}{4}. \] Let $AE=3x$ and $EC=4x$. Then $7x=AE+EC=AC=14$, so $x=2$, and $AE=3x=6$.

答案（E）：因为$\triangle AED$和$\triangle BEC$面积相等，所以$\triangle ACD$和$\triangle BCD$的面积也相等。线段$CD$是$\triangle ACD$与$\triangle BCD$的公共边，因此从$A$与$B$到$CD$的高相等。于是$AB \parallel CD$，所以$\triangle ABE$与$\triangle CDE$相似，相似比为 \[ \frac{AE}{EC}=\frac{AB}{CD}=\frac{9}{12}=\frac{3}{4}. \] 设$AE=3x$，$EC=4x$。则$7x=AE+EC=AC=14$，所以$x=2$，并且$AE=3x=6$。

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