2009 AMC 10B Problem 20

Problem 20 of 25HarderGeometry

Triangle

ABC

has a right angle at

B,

AB=1,

and

BC=2.

The bisector of

\angle BAC

meets

\overline{BC}

D.

What is

BD?

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Solution

By the Pythagorean Theorem,

AC=\sqrt{1^2+2^2}=\sqrt5.

The Angle Bisector Theorem gives

\dfrac{BD}{DC}=\dfrac{AB}{AC}=\dfrac{1}{\sqrt5},

DC=\sqrt5\,BD.

Since

BD+DC=2,

we have

BD(1+\sqrt5)=2,

BD=\dfrac{2}{1+\sqrt5}=\dfrac{\sqrt5-1}{2}.

Thus, the correct answer is B.