Proving Segment Relationships

The Segment Addition Postulate

If $B$ is between $A$ and $C$, then $AB + BC = AC$. The postulate works in both directions: given the two parts you can find the whole, and given the whole and one part you can find the other part.

The word “between” is doing real work — $B$ must lie on segment $\overline{AC}$. If the three points are not collinear with $B$ in the middle, the postulate does not apply.

Properties of segment congruence

Congruence of segments behaves like equality of numbers. Reflexive: $\overline{AB} \cong \overline{AB}$. Symmetric: if $\overline{AB} \cong \overline{CD}$, then $\overline{CD} \cong \overline{AB}$. Transitive: if $\overline{AB} \cong \overline{CD}$ and $\overline{CD} \cong \overline{EF}$, then $\overline{AB} \cong \overline{EF}$.

Remember the translation between the two languages: $\overline{AB} \cong \overline{CD}$ means the segments are congruent, while $AB = CD$ means their lengths are equal. The definition of congruence lets you move between the statements, and proofs often need that move spelled out as its own line.

A standard segment proof

A classic setup: points $A$, $B$, $C$, $D$ sit in order on a line, you are given $AB = CD$, and you must prove $AC = BD$. The trick is to add the shared middle piece $BC$ to both sides, then use segment addition to rename each sum. Most segment proofs are variations of this add-the-shared-piece move.

The tick marks in the figure show the given $AB = CD$; the shared middle piece $\overline{BC}$ is what gets added to both.

Worked examples

Try one yourself

Common questions