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

By the Pythagorean Theorem, $AC=\sqrt5$ . The Angle Bisector Theorem now yields that

$\frac{BC}{1}=\frac{2-BC}{\sqrt5}\\ BC\left(1+\frac{1}{\sqrt5}\right)=\frac{2}{\sqrt5}\\ BC(\sqrt5+1)=2\\ BC=\frac{2}{\sqrt5+1}=\boxed{\frac{\sqrt5-1}{2}}.$

2009 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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2009 AMC 10B Problems/Problem 20

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