Proving the equality case in triangle inequality Background
When plotted on a real number line, it may be deduced that if 
$$a,b,c \in \mathbb{R} $$
$$a < b < c$$
then 
$$\left| {a - c} \right| = \left| {a - b} \right| + \left| {b - c} \right|$$
Problem
But the problem is with the proof. How can the above statement be proven true from the properties of order structure and the definition of absolute value? 

Since both sides of the equality are nonnegative, we can square it:
\begin{align}
|a-c| = |a-b| + |b-c| &\iff |a-c|^2 = (|a-b| + |b-c|)^2\\
&\iff a^2-2ac+c^2 = a^2-2ab+b^2+b^2-2bc+c^2 + 2|a-b||b-c|\\
&\iff 0 = b^2 - ab - bc + ac + |a-b||b-c|\\
&\iff 0 = -(a-b)(b-c) + |a-b||b-c|\\
&\iff  (a-b)(b-c) = |a-b||b-c|\\
&\iff a-b \,\text{ and } \,b-c \text{ have the same sign}\\
&\iff a \ge b \ge c \,\text{ or }\,a \le b \le c 
\end{align}
