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 \[\vec{\alpha} , \vec{\beta} ,\vec{\gamma}\] be the position vectors of the vertices  A, B and C respectively.
Let AD, BEand 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 that is c : b.
Then,
P.V. of D is \[\frac{c \vec{\gamma} + b \vec{\beta}}{c + b}\].
P.V. of E is\[\frac{c \vec{\gamma} + a \vec{\alpha}}{c + a}\] and 
P.V. of F is \[\frac{a \vec{\alpha} + b \vec{\beta}}{a + b}\]
The point dividing AD in the ratio \[b + c : a\] is \[\frac{a \vec{\alpha} + b \vec{\beta} + c \vec{\gamma}}{a + b + c}\]
The point dividing BE in the ratio of \[a + c : b\] is\[\frac{a \vec{\alpha} + b \vec{\beta} + c \vec{\gamma}}{a + b + c}\]
The point dividing CF in the ratio of a + b : c is \[\frac{a \vec{\alpha} + b \vec{\beta} + c \vec{\gamma}}{a + b + c}\]
Since the point \[\frac{a \vec{\alpha} + b \vec{\beta} + c \vec{\gamma}}{a + b + c}\]  lies on all the three internal bisectors AD, BE and CF.
Hence the internal bisectors are concurrent .