In a triangle ABC, DE is a line segment which intersects AB at D and AC at E such that DE is parallel to BC. The line segment divides the triangle in two parts of equal area. What is \frac{BD}{AB} equal to ?
- A. \frac{\sqrt{2}-1}{2}
- B. \frac{\sqrt{2}-1}{\sqrt{2}} ✓
- C. \frac{\sqrt{3}-1}{\sqrt{3}}
- D. \sqrt{2}
Correct Answer: B. \frac{\sqrt{2}-1}{\sqrt{2}}
Explanation
Since DE \parallel BC, \triangle ADE \sim \triangle ABC. The ratio of their areas is (\frac{AD}{AB})^2 = \frac{1}{2} \implies \frac{AD}{AB} = \frac{1}{\sqrt{2}}. Using BD = AB - AD, we find \frac{BD}{AB} = 1 - \frac{AD}{AB} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2}-1}{\sqrt{2}}.
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