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2009 AMC 10B Problems/Problem 20

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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 $BD$ ?

$\text{(A) } \frac {\sqrt3 - 1}{2}\qquad\text{(B) } \frac {\sqrt5 - 1}{2}\qquad\text{(C) } \frac {\sqrt5 + 1}{2}\qquad\text{(D...$

Solution

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

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

See Also

2009 AMC 10B (Problems • Resources)
Preceded by Problem 19	Followed by Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

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