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 $\vec{a},\vec{b},\vec{c}$ represent the sides $A,B,C$ respectively.
The angle bisectors are along
$\dfrac{\vec{a}}{|\vec{a}|}+\dfrac{\vec{b}}{|\vec{b}|},\quad
\dfrac{\vec{b}}{|\vec{b}|}+\dfrac{\vec{c}}{|\vec{c}|},\quad
\dfrac{\vec{c}}{|\vec{c}|}+\dfrac{\vec{a}}{|\vec{a}|}$.
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 $$D=\frac{x\vec{c}+z\vec{b}}{x+z}$$
What should be the next step? Or is there a better method?
Best Answer
You just need to apply the Angle Bisector Theorem again. For completeness, I'll go through the whole argument.
Given $\triangle ABC$ with side-lengths $a = |\overrightarrow{BC}|$, $b = |\overrightarrow{CA}|$, $c := |\overrightarrow{AB}|$, suppose the angle bisector from $A$ meets the opposite side at $D$. You're correct that, by the Angle Bisector Theorem $$\frac{|\overrightarrow{BD}|}{|\overrightarrow{DC}|} = \frac{|\overrightarrow{AB}|}{|\overrightarrow{AC}|} = \frac{c}{b} \tag{$\star$}$$ so that we can write: $$D = \frac{B b + C c}{b+c}$$
Note also that $(\star)$ implies $$\frac{|\overrightarrow{BD}|}{|\overrightarrow{BC}|} = \frac{c}{b+c} \qquad\text{so that}\qquad |\overrightarrow{BD}| = |\overrightarrow{BC}|\frac{c}{b+c} = \frac{ac}{b+c}$$
Now, here's your next step: Suppose the angle bisector from $B$ meets $\overline{AD}$ at $X$ (which we "know" is the incenter). Again by the Angle Bisector Theorem applied to $\triangle ABD$, $$\frac{|\overrightarrow{AX}|}{|\overrightarrow{XD}|} = \frac{|\overrightarrow{BA}|}{|\overrightarrow{BD}|} = \frac{c}{ac/(b+c)} = \frac{b+c}{a}$$ and we can write
Because this formula is symmetric in the elements of $\triangle ABC$, we conclude that $X$ is not merely where the angle bisector from $A$ meets the angle bisector from $B$, but where any two angle bisectors meet ... and, therefore, where all three bisectors meet (aka, the incenter). $\square$