Show, by vector methods, that the angularbisectors of a triangle are concurrent and find an expression for the position vector of the point of concurrency in terms of the position vectors of the vertices.

Explanation

Let `veca, vecb and vecc` be the position vectors of A, B and C respectively.
Let `AD, BE and CF` be the bisectors of `angle A, angle B and angleC`, respectively.
a, b and c are the lengths of sides BC, CA and AB, respectively.
Now AD divides BC in the ratio
`" "BD : DC =AB : AC = c:b`.
Hence, the position vector of D is `vecd = (b vecb + c vecc)/(b+c)`.
Let `I` be the point of intersection of BE and AD.
Then in `Delta ABC, BI` is bisector of `angle B`. Therefore,
`DI : IA =BD : BA`
But `(BD)/(DC) = (c)/(b) or (BD) /(BD + DC) = (c)/(c+b)`

or `(BD)/(BC) = (c)/(c+b)`
or `BD = (ac)/(b+c)`
`therefore DI : IA = (ac)/(b+c) : c = a: (b+c)`
`therefore ` P.V. of `I = (veca a + vecd (b+c))/(a+b+c)`
`" " = (a veca + ((b vecb + c vecc)/(b +c)) (b+c))/(a+b+c)`
`" " = (a veca + b vecb + c vecc)/(a+ b+ c)`
As P.V. of I is symmetrical in `veca, vecb, vecc and a, b, c`, it must lie on CF as well.