Prove that angle bisectors of a triangle are concurrent using vectors. Also, find the position vector of the point of concurrency in terms of position vectors of the vertices.

I solved this without using vectors to get some idea. I am not sure how to prove it using vectors. I don't want to use vector equations for straight lines and then find the point of concurrency. That's like solving using coordinate geometry.

Let $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ represent the sides $A,B,C$ respectively.

The angle bisectors are along $\frac{\overrightarrow{a}}{|\overrightarrow{a}|}}+{\displaystyle \frac{\overrightarrow{b}}{|\overrightarrow{b}|}},\phantom{\rule{1em}{0ex}}{\displaystyle \frac{\overrightarrow{b}}{|\overrightarrow{b}|}}+{\displaystyle \frac{\overrightarrow{c}}{|\overrightarrow{c}|}},\phantom{\rule{1em}{0ex}}{\displaystyle \frac{\overrightarrow{c}}{|\overrightarrow{c}|}}+{\displaystyle \frac{\overrightarrow{a}}{|\overrightarrow{a}|}$

Let the sides $AB,BC,CA$ be $x,y,z$. Let $AD$ be one of the angular bisector.

$\frac{BD}{CD}=\frac{x}{z}$

Hence

$D=\frac{x\overrightarrow{c}+z\overrightarrow{b}}{x+z}$

What should be the next step? Or is there a better method?