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+5​2​=(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​​.
