The vertices A, B, C of triangle ABC have respectively position v… - Vidyadip

Question

The vertices A, B, C of triangle ABC have respectively position vector $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ with respect to given origin O. Show that the point D where the bisector of $\angle{\text{A}}$ meets BC has position Vector $\vec{\text{d}}=\frac{\beta\vec{\text{b}}+\gamma\vec{\text{c}}}{\beta+\gamma}$, where $\beta=\big|\vec{\text{c}}-\vec{\text{a}}\big|$ and, $\gamma=\big|\vec{\text{a}}-\vec{\text{b}}\big|$.

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Answer

Let the position vectors of A, B and C with respect to same origin, O be $\vec{\text{a}},\ \vec{\text{b}}\text{ and }\vec{\text{c}}$ respectively.
Let D be the point on BC where bisectors of $\angle\text{A}$ meets.
Let $\vec{\text{d}}$ be the position vector of D which divides CB internally in the ratio $\beta \text{ and } \gamma$, where
$\beta=\Big|\overrightarrow{\text{AC}}\Big|\text{ and }\gamma=\Big|\overrightarrow{\text{AB}}\Big|$
By section formula, the position vector of D is given by
$\overrightarrow{\text{OD}}=\frac{\beta\vec{\text{b}}+\gamma\vec{\text{c}}}{\beta+\gamma}$
Let $\alpha=\big|\vec{\text{b}}-\vec{\text{c}}\big|$
In center is the concurent point of angle bisectors an in center divides the line AD in the ratio $\alpha : \beta +\gamma$.
So, the position vector of in center is given as,
$\frac{\alpha\vec{\text{a}}+\Big(\frac{\beta\vec{\text{b}}+\gamma\vec{\text{c}}}{\beta+\gamma}\Big)(\beta+\gamma)}{\alpha+\beta+\gamma}=\frac{\alpha\vec{\text{a}}+\beta\vec{\text{b}}+\gamma\vec{\text{c}}}{\alpha+\beta+\gamma}$

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