The points O, A and B are on a plane such that relative to the point O, the points A and B have non-parallel position vectors a and b respectively. The point C with position vector $c$ is on the plane OAB such that OC bisects the angle AOB. The question asks me to show $$\left(\frac{a}{|a|}-\frac{b}{|b|}\right)\cdot c = 0$$ and i can show it easily by using the formula for angles between two vectors.
But next part the question says show that OC is parallel to $\left(\frac{a}{|a|}+\frac{b}{|b|}\right)$
I know this is true as I know the unit vector $a$ plus unit vector $b$ must be in the direction of $c$, but I am not sure how to show it rigorously using theorems.
First we show that if $a$ and $b$ are unit vectors, $a+b$ and $a-b$ are orthogonal to each other. The proof is simple: $(a+b)\cdot(a-b)=a\cdot a-a\cdot b+b\cdot a-b\cdot b=\Vert a\Vert^2-\Vert b\Vert^2=1-1=0$.
The first part of the question gives that $c$ is perpendicular to $\frac a{|a|}-\frac b{|b|}$, which is itself perpendicular to $\frac a{|a|}+\frac b{|b|}$ by the lemma above. Thus, since we are working in a plane, $c$ is parallel to $\frac a{|a|}+\frac b{|b|}$ as required.