Angle bisector/Distance condition/Fact/Proof

We may assume that ${\displaystyle {}v}$ and ${\displaystyle {}w}$ are normed. Let ${\displaystyle {}P\in \mathbb {R} ^{2}}$. Due to fact, we have

${\displaystyle {}d(P,\mathbb {R} v)^{2}=\Vert {P}\Vert ^{2}-\left\langle P,v\right\rangle ^{2}\,,}$

and, accordingly,

${\displaystyle {}d(P,\mathbb {R} w)^{2}=\Vert {P}\Vert ^{2}-\left\langle P,w\right\rangle ^{2}\,.}$

Therefore, the distances coincide if and only if

${\displaystyle {}\left\langle P,v\right\rangle =\pm \left\langle P,w\right\rangle \,}$

holds. If

${\displaystyle {}P=s(v+w)\,,}$

then

${\displaystyle {}\left\langle P,v\right\rangle =\left\langle s(v+w),v\right\rangle =s+s\left\langle w,v\right\rangle =\left\langle s(v+w),w\right\rangle =\left\langle P,w\right\rangle \,,}$

and the equation holds. For the reverse statement, we may write

${\displaystyle {}P=sv+tw\,.}$

From

${\displaystyle {}\left\langle P,v\right\rangle =\left\langle P,w\right\rangle \,}$

we can deduce

${\displaystyle {}s+t\left\langle v,w\right\rangle =\left\langle sv+tw,v\right\rangle =\left\langle sv+tw,w\right\rangle =t+s\left\langle v,w\right\rangle \,;}$

therefore,

${\displaystyle {}s{\left(1-\left\langle v,w\right\rangle \right)}=t{\left(1-\left\langle v,w\right\rangle \right)}\,.}$

Because ${\displaystyle {}v}$ and ${\displaystyle {}w}$ are normed and linearly independent, exercise implies that

${\displaystyle {}\vert {\left\langle v,w\right\rangle }\vert <1\,,}$

hence, the right-hand factor is not ${\displaystyle {}0}$, and, therefore, ${\displaystyle {}s=t}$. In case

${\displaystyle {}\left\langle P,v\right\rangle =-\left\langle P,w\right\rangle \,,}$

a similar consideration gives ${\displaystyle {}s=-t}$.