Let R1 be the vector between origin and P1. Let R2 similar to R1.
Since speed is constant, the p-2 norm of v1 and V2 are both the same. It is trivial that p-2 norm for R1 and R2 are the same, but p-2 norm V1 isn't equal to p-2 norm R1.
Note that R1'*V1 = 0 (cuz they are perpendicular).
Let following linear system be:
| [R1'R1 | R1'R2] | [V1] | = | 0 |
|---|---|---|---|---|
| [R2'R1 | R2'R2] | [V2] | = | 0 |
Note that :
R1'R1 = p^2 (where 'p' is p-2 norm of R1, which is the same for p-2 norm of R2);
R1'R2 = p^2*c (where 'c' is the cossine of the angle between R1 and R2);
moreover, note that R1'R2 denotate dot product between R1'R2, which is equal to p-2 norm R1 * p-2 norm R2 * cossine between both vectors.
The above mentioned linear system can be represented (equivalently) as:
| [1 | c] | [V1] | = | 0 |
|---|---|---|---|---|
| [c | 1] | [V2] | = | 0 |
note that A matrix (coefficients matrix) isn't singular (and hence invertable) if 'c' is different of -1 or 1, which represent the case that R1 and R2 are coincidents (so V1 and V2 would be coincident as well, therefore there is no triangle).
By the previous analysis on A matrix, is possible to assure that V1'V2 has the same angle as R1'R2, ans since both p-2 norm of both Ri and Vj are the same, both triangles are similar (side-angle-side).