# 2009 AMC 10B Problems/Problem 20

## Problem

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 $BD$? $[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); dotfactor=4; pair A=(0,1), B=(0,0), C=(2,0); pair D=extension(A,bisectorpoint(B,A,C),B,C); pair[] ds={A,B,C,D}; dot(ds); draw(A--B--C--A--D); label("1",midpoint(A--B),W); label("B",B,SW); label("D",D,S); label("C",C,SE); label("A",A,NW); draw(rightanglemark(C,B,A,2)); [/asy]$ $\text{(A) } \frac {\sqrt3 - 1}{2} \qquad \text{(B) } \frac {\sqrt5 - 1}{2} \qquad \text{(C) } \frac {\sqrt5 + 1}{2} \qquad \text{(D) } \frac {\sqrt6 + \sqrt2}{2} \qquad \text{(E) } 2\sqrt 3 - 1$

## Solution 1

By the Pythagorean Theorem, $AC=\sqrt5$. Then, from the Angle Bisector Theorem, we have: $\frac{BD}{1}=\frac{2-BD}{\sqrt5}\\ BD\left(1+\frac{1}{\sqrt5}\right)=\frac{2}{\sqrt5}\\ BD(\sqrt5+1)=2\\ BD=\frac{2}{\sqrt5+1}=\boxed{\frac{\sqrt5-1}{2} \implies B}.$

## Solution 2

Let $\theta = \angle BAD = \angle DAC$. Notice $\tan \theta = BD$ and $\tan 2 \theta = 2$. By the double angle identity, $$2 = \frac{2 BD}{1 - BD^2} \implies BD = \boxed{\frac{\sqrt5 - 1}{2} \implies B}.$$

## Solution 3

Let $BD=y$.

Make $DE$ a line so that it is perpendicular to $AC$. Since $AD$ is an angle bisector, $\triangle AED$ is congruent to $\triangle ABD$. Using the Pythagorean Theorem: $AC^2=1^2+2^2$ $AC^2=5$ $AC=\sqrt{5}$

We know that $AE=1$ by the congruent triangles, so $EC=\sqrt{5}-1$. We know that $DE=y$, $EC=\sqrt{5}-1$, and $DC=2-y$. We now have right triangle $DEC$ and its 3 sides. Using the Pythagorean Thereom, we get: $y^2+(\sqrt{5}-1)^2=(2-y)^2$ $-4y=2-2\sqrt{5}$

So, $y=BD=\boxed{\frac{\sqrt5-1}{2} \implies B}.$

~HelloWorld21

## Video Solution

~ pi_is_3.14

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