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? 

To prove it you do cases
$|a-c| = \pm (a-c)$ and $|a-b| + |b- c| = \pm(a-b) + \pm (b-c)$.
consider the ways that $a,b,c$ may be ordered:
(your case is a subcase of case 6: so you can skip to the very last case if you want.)
1:  Assume $a \ge b \ge c$ then
$|a-c| = a-c$ and $|a-b| + |b-c| = (a-b) + (b-c) = a-c = |a-c|$.
2:  Assume $a \ge c \ge b$ then 
$|a-c| = a-c$ and $|a-b| + |b-c| = (a-b) -(b-c) = a-2b + c$.  Now $2b \le 2c$ so $-2b \ge -2c$ and $a-2b + c\ge a -2c +c = a-c = |a-c|$.
3: Assume $b \ge a \ge c$ then
$|a-c| = a-c$ and $|a-b| + |b-c| = -(a-b) +(b-c) = -a+2b - c$.  Now $2b \ge 2a$ so  $-a+2b - c \ge  -a +2a -c = a-c = |a-c|$.
4: Assume $b \ge c \ge a$
$|a-c| = -(a-c) = c-a$ and $|a-b| + |b-c| = -(a-b) +(b-c) = -a+2b - c$.  Now $2b \ge 2c$ so  $-a+2b - c \ge  -a +2c -c = -a+c = |a-c|$.
5: Assume $c \ge a \ge b$
$|a-c| = -(a-c) = c-a$ and $|a-b| + |b-c| = (a-b) -(b-c) = a-2b + c$.  Now $2b \le 2a$ so $-2b \ge -2a$ and $a-2b + c\ge a -2a +c = -a+c = |a-c|$.
6: Assume $c \ge b \ge a$
$|a-c| = -(a-c) = c-a$ and $|a-b| + |b-c| = -(a-b) - (b-c) = (b-a) + (c-b) = -a + c = |a-c|$.
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The "equality cases are 1) and 6) and they are very straightforward.
