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2009 AMC 10B Problem 20

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2009 AMC 10B Problem 20AMC 10 Step by Step
Name:                             Date: September 13, 2026
Triangle Centers & Cevians · difficulty 3 of 5 · about 4 min · position P16-20

Triangle ABCABC has a right angle at BB , AB=1AB=1 , and BC=2BC=2 . The angle bisector of ∠A\angle A intersects side BC‾\overline{BC} at DD . What is BDBD ?

figure

  1. A)

    3−12\frac {\sqrt3 - 1}{2}

  2. B)

    5−12\frac {\sqrt5 - 1}{2}

  3. C)

    5+12\frac {\sqrt5 + 1}{2}

  4. D)

    6+22\frac {\sqrt6 + \sqrt2}{2}

  5. E)

    23−12\sqrt 3 - 1

Working