Geometric proof - distances on a straight line

Question

How would you prove that given 3 points $A, B, C \in \mathbb{R}^2 $, $|AC|=|AB|+|BC| $, (where $B$ is between $A$ and $C$) if and only if the points $A, B$ and $C$ are collinear.

Answer 1

If $A$, $B$, $C$ are collinear than it automatically implies $|AB|+|BC|=|AC|$.

To prove the converse:

Let $\theta = \angle ABC$, $|AB|=p$, $|BC|=q$ and $CA=r$. So, by the cosine rule, $$r=\sqrt{p^2+q^2-2pq\cos \theta}$$ We know $r=p+q$, hence $$p+q=\sqrt{p^2+q^2-2pq\cos \theta}$$ Squaring, $$p^2+q^2+2pq=p^2+q^2-2pq\cos \theta$$ Which implies, $$\cos \theta = -1$$ and since $0 \leq \theta \leq 2\pi$ we deduce $\theta = \angle ABC=\pi$. Hence $ABC$ is a straight line.

[Note: This proof is not recursive because the parallelogram law of vectors doesn't require such collinearity in its derivation.]

[Note 2: I am a student, new to math.stackexchange and have English as my second language so please do correct me if I am wrong anywhere.]