If the interior and exterior bisectors of the angle $A$ of a triangle $ABC$ meet the side $BC$ at $D$ and $E$, then :
Step-by-Step Solution
Key Concept: By the angle bisector theorem, the internal bisector divides BC in ratio AB:AC (giving BD = ac/(b+c)), while the external bisector divides BC externally in the same ratio (giving BE = ac/(c-b)). The harmonic relationship 2/BC = 1/BD + 1/BE emerges from these ratios.
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