A segment having equal angles with two equal segments is perpendicular to the connecting line Let $a,b,c \in \mathbb{R}^3$ be unit vectors, and suppose that the angles between $a,b$, and between $a,c$ are equal.
Is there an elementary, geometric, computation-free proof that $a$ is perpendicular to $b-c$?
i.e. the claim is that a segment having equal angles with two segments of equal length is perpendicular to the line connecting them.
I want a proof that do not use the notion of inner product. After all, the notion of orthogonality predates the inner product. (At least pedagogically, I guess that historically as well).
Of course, if you use inner products, the proof is trivial:
$$
\langle a,b \rangle=|a||b|\cos \theta_{a,b}=\cos \theta_{a,b}=\cos \theta_{a,c}=\langle a,c \rangle,
$$
so $\langle a,b-c \rangle=0.$
The best would be an elementary proof you can present to high-school students:) A picture here is very convincing, but I am looking for a rigorous proof.

Comment: If $a,b,c$ lie on a plane, this is the claim that angle bisector in isosceles triangle is perpendicular to the base.
 A: Let us then consider the case when the vectors are as in the following picture:

We need only the fact that the vectors
$\overrightarrow b=\overrightarrow {OB}$ and
$\overrightarrow c=\overrightarrow {OC}$ have the same length, so
$OB=OC$ for short. Let $H$ be the projection of $A$ on the plane $(OBC)$.
Let $OH$ intersect $BC$ in a point $K$.
The question gives $\widehat{AOB}=\widehat{AOC}$, so the triangles
$\Delta AOB$ and $\Delta AOC$ are congruent. (The side $AO$ is common, and we have the same angle, the same sides in $O$.) We get $AB=AC$.
We compare $\Delta AHB$ and $\Delta AHC$ now. Congruent, $AH$ is common, we have right angles in $H$, and the hypotenuses are congruent. So $HB=HC$, i.e. $H$ is as $O$ on the side bisector of $BC$, so denoting by $K$ the mid point of the segment $BC$ we have:
$$
OHK\perp BC\ .
$$
The angle between $\overrightarrow a:=\overrightarrow {OA}$ and
$\overrightarrow b-\overrightarrow c=\overrightarrow {CB}$ is then by definition the angle between the line $OA$ and a parallel to $BC$ through $O$. Note that this parallel is perpendicular on the plane
$(AOHK)$ - since it is on the two lines $AH$ and $OHK$ generating it - so it is also perpendicular on the line $OA$ of this plane.
$\square$
