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, $\gamma=\big|\vec{\text{a}}-\vec{\text{b}}\big|$.