Show by vector methods that the angular bisectors of a triangle are concurrent and find the position of the point of concurrency in terms of the position vectors of the vertices.
Let the vertices of the triangle by and . We use the geometrical fact that an angle bisector divides the opposite side in the ratio of the sides containing the angle.
We thus have,
Thus, is given by (the internal division formula):
In , since is the angle bisector, we have
Thus, we now have the position vectors of and we know what ratio divides in. can now be easily determined using the internal division formula:
The symmetrical nature of this expression proves that the bisector of will also pass through . The angle bisectors will therefore be concurrent at , called the incentre. The position vector of the incentre is given by
In a parallelogram , let be the mid-point of . and meet in . Prove that both and are trisected at .
There’s no loss of generality in assuming to be the origin . Let and have the position vectors and . The position vector of is given by where
is therefore the point .
Since is the mid-point of , ‘s position vector is . Let us first find the position vector of a point ‘ which lies on and trisects it. We’ll then show that same point lies on and trisects on and trisects it too, proving the stated assertion.
A point ” which trisects is
Since ‘ and ” are the same-point, say , we see that and are trisected at .