Vector Algebra
Vectors
star_batch_jee_advanced_2025
Grade 12
Question:
If the interior and exterior bisectors of the angle $A$ of a triangle $ABC$ meet the side $BC$ at $D$ and $E$, then :
$2BC = BD + BE$
$BC^2 = BD \cdot BE$
$\frac{2}{BC} = \frac{1}{BD} + \frac{1}{BE}$
None of these
Step-by-Step Solution
Key Concept: The angle bisector theorem relates the segments created on the opposite side in terms of adjacent side lengths.
The internal bisector from $A$ is $\vec{r} = \vec{a} + t\left(\frac{\vec{b}}{c} + \frac{\vec{c}}{b}\right)$. The line $BC$ has equation $\vec{r} = \vec{b} + k(\vec{c} - \vec{b})$. The point $D$ where the bisector meets $BC$ satisfies $BD = \frac{c|\vec{c} - \vec{b}|}{b + c}$ and $BE = \frac{c|\vec{c} - \vec{b}|}{c - b}$, yielding $\frac{1}{BE} + \frac{1}{BD} = \frac{2}{BC}$.
Correct Answer: 3