Equality from distance of point in segment in inner product norm

Question

For some reason I can't seem to get this simple result:
$c = ta+(1-t)b$, where $a,b\in \mathbb{R}^n$ , iff $||b-a|| = ||b-c||+||c-a|| $
Where $||\cdot||$ comes from an inner product. Edit: I see now how to prove the forward direction, but I'm having difficulty with the converse...

Answer 1

For the converse, take the equation $||b-a||-||c-b||=||c-a||$ and square it for the result:

$||b||^2-a\cdot b-b\cdot c+a\cdot c=||b-a||\hspace{0.1cm}||c-b||$

which can be conveniently rewritten as:

$(b-c)\cdot(b-a)=||b-c||\hspace{0.15cm}||b-a||$ for all vectors a,b,c.

However the Schwarz inequality tells us that:

$(b-c)\cdot(b-a)\leq||b-c||\hspace{0.15cm}||b-a||$ with the equality being true iff $b-c=t(b-a), t\in \mathbb{R}$. By rearranging the latter equation the proof of the converse is complete.