The vertices A, B, C of triangle ABC have respectively position vectors with respect to a given origin O. Show that the point D where the bisector of ∠A meets BC has position vector where and

Given the position vectors of vertices A, B and C of ΔABC are and respectively.

D is point on BC with position vector such that AD is the bisector of ∠A. I is the incenter of ΔABC.

Observe from the figure that D divides BC in the ratio BD:DC.

Using the angular bisector theorem, we know that the angle bisector of an angle in a triangle bisects the opposite side in the ratio equal to the ratio of the other two sides.

But, and.

Recall the vector is given by

Similarly,

So, we have .

Recall the position vector of point P which divides AB, the line joining points A and B with position vectors and respectively, internally in the ratio m : n is

Here, we have D dividing BC internally in the ratio m:n where m = BD = and n = DC =

Suppose and.

From angular bisector theorem above, we have.

Adding 1 to both sides,

In addition, as CI is the angular bisector of ∠C in ΔACD, using the angular bisector theorem, we have

So, we get

We have and

Assume

So, I divides AD in the ratio (β + γ):α.

Let the position vector of I be .

Using the aforementioned section formula, we can write

But, we already found.

Thus, and the position vector of the incenter is , where, and .