Question

The vertices A, B, C of triangle ABC have respectively position vectors \[\vec{a}\], \[\vec{b}\], \[\vec{c}\]  with respect to a given origin O. Show that the point D where the bisector of ∠ A meets BC has position vector \[\vec{d} = \frac{\beta \vec{b} + \gamma \vec{c}}{\beta + \gamma},\text{ where }\beta = \left| \vec{c} - \vec{a} \right| \text{ and, }\gamma = \left| \vec{a} - \vec{b} \right|\]
Hence, deduce that the incentre I has position vector
\[\frac{\alpha \vec{a} + \beta \vec{b} + \gamma \vec{c}}{\alpha + \beta + \gamma},\text{ where }\alpha = \left| \vec{b} - \vec{c} \right|\]

Answer 1

Let the position vectors of A, B and C with respect to some origin, O be \[\vec{a} , \vec{b}\text{ and }\vec{c}\] respectively.
Let D be the point on BC where bisectors of ∠A meets.
Let \[\vec{d}\]  be the position vector of D which divides CB internally in the ratio β and γ, where \[\beta = \left| \vec{AC} \right|\text{ and }\gamma = \left| \vec{AB} \right|\]
Thus,
\[\beta = \left| \vec{c} - \vec{a} \right|\text{ and }\gamma = \left| \vec{b} - \vec{a} \right|\] By section formula, the position vector of D is given by

\[\vec{OD} = \frac{\beta \vec{b} + \gamma \vec{c}}{\beta + \gamma}\]
\[\text { Let } \alpha = \left| \vec{b} - \vec{c} \right|\]
Incentre is the concurrent point of angle bisectors and incentre divides the line AD in the ratio ∝: β + γ.
So, the position vector of incentre is given as,
\[\frac{\alpha \vec{a} + \left( \frac{\beta \vec{b} + \gamma \vec{c}}{\beta + \gamma} \right) \left( \beta + \gamma \right)}{\alpha + \beta + \gamma} = \frac{\alpha \vec{a} + \beta \vec{b} + \gamma \vec{c}}{\alpha + \beta + \gamma}\]