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Let $a,b,c$ be three different points in a vector space $E$ over the real numbers, and with an inner product. Let $d$ be the metric defined by the inner product. Prove that if: $ d(a,c) = d(a,b)+d(b,c) $ then $ c= tb + (1-t)a $ for some $ t > 1 $.

The only thing I could do , is write the equality in terms of the dot product, but I don´t know how to use the new equality :/

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Let $u = b-a, v = c-b$. Then you have to prove that if $\Vert u+v\rVert = \lVert u\rVert + \lVert v\rVert$ then $v = tu$ for some $t > 0$.

By explaining the equality in terms of inner product and raising both sides to the power of two, you'll get $$\langle u, u\rangle + 2\langle u, v\rangle + \langle v, v\rangle = \langle u, u\rangle + 2\sqrt{\langle u, u\rangle*\langle v, v\rangle } + \langle v, v\rangle,$$ from which follows $$\langle u, v\rangle^2 = \langle u, u\rangle*\langle v, v\rangle,$$ where $\langle u, v\rangle$ is positive , from which, by inner product linearity, follows $\langle e_u, e_v\rangle = 1$ (where $e_u = \frac{u}{\lVert u\rVert}, e_v = \frac{v}{\lVert v\rVert}$).

Then it is obvious that $$\langle e_u-e_v, e_u-e_v\rangle = \langle e_u, e_u\rangle - 2\langle e_u, e_v\rangle+ \langle e_v, e_v\rangle = 0,$$ meaning that $e_u = e_v = e$, $u = e*\lVert u\rVert, v = e*\lVert v\rVert$, $v = u*\frac{\lVert v\rVert}{\lVert u\rVert}$, QED.

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