Problem:
Triangle ABC has a right angle at B,AB=1, and BC=2. The bisector of ∠BAC meets BC at D. What is BD?
Answer Choices:
A. 23−1
B. 25−1
C. 25+1
D. 26+2
E. 23−1
Solution:
By the Pythagorean Theorem, AC=5. By the Angle Bisector Theorem, ABBD=ACCD. Therefore CD=5⋅BD and BD+CD=2, from which
BD=1+52=(B)25−1
OR
Let DE be an altitude of △ADC. Then note that △ABD is congruent to △AED, and so AE=1. As in the first solution AC=5. Let x=BD. Then DE=x,EC=5−1, and DC=2−x. Applying the Pythagorean Theorem to △DEC yields x2+(5−1)2=(2−x)2, from which x=(B)25−1.
