The Segment Addition Postulate

Two postulates about segments are so basic that they need no proof of their own — yet almost every geometric proof that mentions lengths leans on them. Together they let you break a segment into parts, or rebuild it from parts, whenever convenient.

Concept The two postulates

Segment length postulate — every segment has a unique positive length, a real number.

Segment addition postulate — if

is between

and

on the same line, then

The whole segment

is made of exactly the two smaller pieces

and

, so their lengths must add to the length of the whole.

Postulates are accepted as true without proof — they are the starting assumptions on which every later theorem is built.

Example Finding a missing length

Point lies between and . If and , find .

By the segment addition postulate:

Substitute:

Solve:

⟹ BC = 38

Example Proof with intersecting segments

Given: lies between and , and between and , with and . Prove that .

Statement	Reason
CE ≅ FE, ED ≅ EG	Given
CE = FE, ED = EG	Definition of congruent segments
CE + ED = CD, FE + EG = FG	Segment addition postulate
CD = FG	Substitution
CD ≅ FG	Definition of congruent segments

The proof moves back and forth between congruence (≅) and equality (=): congruent segments have equal lengths, so switching to lengths lets ordinary algebra do the rest of the work.

Example Proof with four points on a line

Given: points , , , lie on the same line in that order, and . Prove that .

From the given:

By the segment addition postulate:

and

Substitute:

Subtract

from both sides:

⟹ PQ ≅ RS

The shared middle segment appears in both sums, so subtracting it from each side is exactly what isolates the two segments being compared — a pattern worth recognising whenever four collinear points appear together.

Summary

Every segment has a unique, positive real-number length.
If B is between A and C, then AB + BC = AC — the segment addition postulate.
Proofs convert between ≅ and = using the definition of congruent segments, then use substitution and subtraction to finish.
When a shared middle segment appears in two sums, subtracting it out is often the key final step.