AMC10 2009 B

AMC10 2009 B · Q20

AMC10 2009 B · Q20. It mainly tests Word problems (algebra), Triangles (properties).

Triangle $ABC$ has a right angle at $B$, $AB = 1$, and $BC = 2$. The bisector of $\angle BAC$ meets $\overline{BC}$ at $D$. What is $BD$?

$\triangle ABC$ 在 B 处直角，$AB = 1$，$BC = 2$。$\angle BAC$ 的平分线与 $\overline{BC}$ 相交于 D。求 $BD$？

(A) $\frac{\sqrt{3}-1}{2}$ $\frac{\sqrt{3}-1}{2}$

(B) $\frac{\sqrt{5}-1}{2}$ $\frac{\sqrt{5}-1}{2}$

(D) $\frac{\sqrt{6}+\sqrt{2}}{2}$ $\frac{\sqrt{6}+\sqrt{2}}{2}$

(E) $2\sqrt{3}-1$ $2\sqrt{3}-1$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): By the Pythagorean Theorem, $AC=\sqrt{5}$. By the Angle Bisector Theorem, $\frac{BD}{AB}=\frac{CD}{AC}$. Therefore $CD=\sqrt{5}\cdot BD$ and $BD+CD=2$, from which $$ BD=\frac{2}{1+\sqrt{5}}=\frac{\sqrt{5}-1}{2}. $$

答案（B）：由勾股定理，$AC=\sqrt{5}$。由角平分线定理，$\frac{BD}{AB}=\frac{CD}{AC}$。因此 $CD=\sqrt{5}\cdot BD$ 且 $BD+CD=2$，从而 $$ BD=\frac{2}{1+\sqrt{5}}=\frac{\sqrt{5}-1}{2}. $$

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