# Segment Addition Postulate: Definition, Formula, Examples, FAQs

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## What Is the Segment Addition Postulate in Geometry?

The segment addition postulate states that if three points A, B, and C are collinear such that B lies between A and C, then the sum of the lengths of segment AB and segment BC is equal to the length of the entire segment AC.

The segment addition postulate (segment addition theorem), in simple words, states that if we divide a line segment into smaller segments, the sum of lengths of the smaller segments will add up to the length of the original segment. It is applicable to the line segments.

The segment addition postulate is an important property of line segments that is used to check if three points are collinear or whether a point lies on a given segment or not.

The Segment Addition Postulate is a fundamental principle in geometry that states that if three points A, B, and C are collinear such that B lies somewhere on AC, then the sum of the lengths of the segments AB and BC is equal to the length of the entire segment AC.

The formula for the segment addition postulate with respect to three collinear points A, B, C is given by

l(AB) + l(BC) = l(AC)

OR

AB + BC = AC

## How to Use ​​Segment Addition Postulate

Let’s understand the steps to use the segment addition postulate with the help of an example.

Example: Find the length of the line segment AB.

Step 1: Identify the collinear points and note down the given lengths of the line segments.

Here, C lies between A and B.

AC = 4

BC = 7

Step 2: Write the segment addition formula with respect to the given collinear points.

Here, AC + CB = AB

Step 3: Substitute the values and simplify.

AB = 4 + 8 = 12 units

## How to Know if Three Points are Collinear

• If the point P lies between A and B, then AP + PB = AB.
• If AP + PB = AB, then P lies between A and B, and the points A, P, and B are collinear.

• The segment addition postulate is not applicable to lines or rays.
• This postulate can identify the midpoint of the line segment. If AB + BC = AC and AB = BC, then B is the midpoint of AC.

## Conclusion

In this article, we learned about the segment addition postulate, which is a fundamental property of line segments that helps us to identify collinear points. Let’s solve a few examples and MCQs for better understanding.

## Solved Examples on Segment Addition Postulate

1. Does point B lie on segment AC if segment AB = 3 units, BC = 5 units, and AC = 6 units?

Solution:

AB = 3 units, BC = 5 units, and AC = 6 units

Let us calculate the sum of AB and BC

l(AB) + l(BC) = 3 + 5 = 8 units

l(AC)=6 units

l(AB) + l(BC)$/neq$ l(AC)

Thus, B does not lie on the line segment AC.

2. In the given diagram, l(AC) = 28 units. Find x.

Solution:

B lies between the points A and C.

AB + BC = AC

2x + 3x + 3 = 28

5x + 3 = 28

5x = 25

x = 5

Thus, the value of x is 5.

3. The point R lies on the line segment PS, where PR = 5x, RS = (6x + 1) , and PS = 56 units. What will be the value of x, PR, RS and PS

Solution:

As point S lies on the line segment PS, the equation will be:

PR+RS=PS

5x+6x+1=56

11x + 1 = 56

11x = 56 – 1

11x = 55

$x = \frac{55}{11}$

x =5

PR = 5$\times$ 5 = 25 units

RS = 6(5) + 1=31 units

4. Find GH.

Solution:

G lies between F and H.

FH = 35 units

GF = 20 units

By segment addition postulate, FG + GH = FH

GH = FH – FG

GH = 35 – 20

GH = 15 units

## Practice Problems on Segment Addition Postulate

1

1
13
12
14
CorrectIncorrect
RS + ST = RT
1 + ST = 13
ST = 12
2

### If the point Y bisects the line segment XZ, then

XY = YZ
XZ = 2XY
XZ = 2YZ
All of the above
CorrectIncorrect
Correct answer is: All of the above
By the segment addition postulate, XY + YZ = XZ
Since Y is midpoint of XZ, we have XY = YZ
Thus, XZ = 2XY = 2YZ
3

### The points IJKL are collinear: IJ = 9, JK = 11, and IL = 26. What is the length of KL?

3
4
5
6
CorrectIncorrect
IJ + JK + KL = IL
9 + 11 + KL = 26
KL = 6 units
4

### What will be the value of x if Q lies on line segment PR such that PQ = 3x - 3, QR = 2x-1, and PR = 11?

8
7
3
10
CorrectIncorrect
$PQ + QR = PR$
$3x - 3 + 2x-1 = 11$
$5x = 11 + 4$
$x = 3$