AMC10 2003 A

AMC10 2003 A · Q22

AMC10 2003 A · Q22. It mainly tests Similarity, Coordinate geometry.

In rectangle ABCD, we have AB = 8, BC = 9, H is on BC with BH = 6, E is on AD with DE = 4, line EC intersects line AH at G, and F is on line AD with GF $\perp$ AF. Find the length GF.

在矩形 ABCD 中，AB = 8，BC = 9，H 在 BC 上且 BH = 6，E 在 AD 上且 DE = 4，直线 EC 与直线 AH 相交于 G，F 在直线 AD 上且 GF $\perp$ AF。求 GF 的长度。

(A) 16 16

(B) 20 20

(D) 28 28

(E) 30 30

Answer

Correct choice: (B)

正确答案：（B）

Solution

(B) We have $EA = 5$ and $CH = 3$. Triangles $GCH$ and $GEA$ are similar, so \[ \frac{GC}{GE}=\frac{3}{5} \quad\text{and}\quad \frac{CE}{GE}=\frac{GE-GC}{GE}=1-\frac{3}{5}=\frac{2}{5}. \] Triangles $GFE$ and $CDE$ are similar, so \[ \frac{GF}{8}=\frac{CE}{GE}=\frac{5}{2} \] and $FG=20$.

（B）我们有 $EA = 5$ 且 $CH = 3$。三角形 $GCH$ 与 $GEA$ 相似，所以 \[ \frac{GC}{GE}=\frac{3}{5} \quad\text{并且}\quad \frac{CE}{GE}=\frac{GE-GC}{GE}=1-\frac{3}{5}=\frac{2}{5}. \] 三角形 $GFE$ 与 $CDE$ 相似，所以 \[ \frac{GF}{8}=\frac{CE}{GE}=\frac{5}{2} \] 且 $FG=20$。

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