Question

Prove by vector method that the Internal bisectors of the angles of a triangle are concurrent .

Answer 1

Let ABC be a triangle and $\overrightarrow{\alpha },\overrightarrow{\beta },\overrightarrow{\gamma }$ be the p.v.s of the vertices A, B and C respectively.

Let AD, BE and CF be the internal bisectors of $\angle A,\angle B$ and $\angle C$ respectively.

We know that D divides BC in the ratio of AB$:$AC i.e., c$:$b.

p.v of D is $\frac{c\overrightarrow{\gamma }+b\overrightarrow{\beta }}{c+a}$

p.v of E is $\frac{c\overrightarrow{\gamma }+a\overrightarrow{\alpha }}{c+a}$

p.v of F is $\frac{a\overrightarrow{\alpha }+b\overrightarrow{\beta }}{a+b}$

The point dividing AD in the ratio $b+c:a$ is $\frac{a\overrightarrow{\alpha }+b\overrightarrow{\beta }+c\overrightarrow{\gamma }}{a+b+c}$

The point of dividing BE in the ratio of $a+c:b$ is $\frac{a\overrightarrow{\alpha }+b\overrightarrow{\beta }+c\overrightarrow{\gamma }}{a+b+c}$

The point dividing CF in the ratio of $a+b:c$ is $\frac{a\overrightarrow{\alpha }+b\overrightarrow{\beta }+c\overrightarrow{\gamma }}{a+b+c}$

Since the point $\frac{a\overrightarrow{\alpha }+b\overrightarrow{\beta }+c\overrightarrow{\gamma }}{a+b+c}$ lies on all the three internal bisectors AD, BE and CF.

Hence, the internal bisectors and concurrent.