Segments

Segment Addition
Segment addition problems can sometimes seem confusing, but as long as you remember that the two smaller parts make up the whole, you will be in good shape!

This means:
one small part + other small part = whole segment

Segment Addition
one small part + other small part = whole segment
A
B
C
AB
BC
AC
AB + BC = AC
This is the Segment Addition Postulate

Segment Addition
A
B
C
AB
BC
AC
AB + BC = AC
Example 1: If AB = 23 and BC = 47, find AC

Segment Addition
A
B
C
23
47
AC
AB + BC = AC
Example 1: If AB = 23 and BC = 47, find AC
23
+ 47 = AC 70 = AC

Segment Addition
A
B
C
4x + 5
15
60
AB + BC = AC
Example 2: If AB = 4x + 5, BC = 15, and AC = 60 find x.
4x + 5
+ 15 = 60
4x + 20 = 60 4x = 40 x = 10

Segment Addition
A
B
C
28
?
58
AB + BC = AC
Example 3: If AB = 28 and AC = 58 find BC.
2828
+ BC = 58+ x = 58
x = 30Which means BC = 30

+ - x + 24 = 5x - 12x + 16 = 5x 1216 = 4x 12
Segment Addition
A
B
C
2x - 8
-x + 24
5x - 12
AB + BC = AC
Example 4: If AB = 2x 8, BC = - x + 24, and AC = 5x 12, find AC.
2x - 8
28 = 4x7 = x
Now that we know x = 7 we can easily find AC by replacing 7 with x in 5x 12. 5x 12 = AC5(7) 12 = AC35 12 = AC23 units = AC

Midpoint of a Segment
Midpoint:

We have learned that if A, B, and C are collinear then AB + BC = AC

If AB = BC then B is called the midpoint of AC,

and we can put in the tic marks to show that AB and BC are congruent
A
C
B
A
C
B

Midpoint of a Segment
A
C
B
AB + BC = AC
Example 1: B is the midpoint of AC. If AC = 115 and AB = 5x 10, find x.
AB + BC = AC5x 10 + BC = 1155x 10 + 5x 10 = 11510x 20 = 11510x = 135x = 13.5

Midpoint of a Segment
A
C
B
Example 2: B is the midpoint of AC. Find x, AB, BC, and AC.
4x + 12
5x - 3
AB = BC 4x + 12 = 5x 3 4x + 15 = 5x 15 = x

Midpoint of a Segment
A
C
B
Example 2: B is the midpoint of AC. Find x, AB, BC, and AC.
4x + 12
5x - 3
AB = BC 4x + 12 = 5x 3 4x + 15 = 5x 15 = x
4x + 12 = AB 4(15) + 12 = AB 60 + 12 = AB 72 = AB 72 = BC
AB + BC = AC72 + 72 = AC144 = AC

Segment Bisector
Biplane has TWO wings that are the same size.
Bicycle has TWO wheels that are the same size.
So what do a biplane and a bicycle have in common with a segment bisector?

Segment Bisector
A segment Bisector ensures the segment has been divided into TWO parts that are the same size. The bisector goes through the segment midpoint and we know that:
A
B
C
m
Line m is the segment bisectorB is the midpoint of AC
The marks indicate that each part is congruent

Perpendicular Segment Bisector
A perpendicular segment Bisector is a segment bisector that runs perpendicular to the segment and passes through the midpoint of the segment.
A
B
C
m
Line m is the perpendicular segment bisectorB is the midpoint of AC
The marks indicate that each part is congruent

Segment Addition with Bisectors
A
B
C
2x - 4
50
Example 4: If m is a bisector of AC, AB = 2x 4, AC = 50, find BC.
BC
m
Recall: AB + BC = AC
AB + BC = AC2x 4 + BC = 50

Segment Addition with Bisectors
A
B
C
2x - 4
50
Example 4: If m is a bisector of AC, AB = 2x 4, AC = 50, find BC.
BC
m
Recall: AB + BC = AC
AB + BC = AC2x 4 + BC = 502x 4 + 2x 4 = 504x 8 = 504x = 58x = 14.5

Segment Addition with Bisectors
A
B
C
2x - 4
50
Example 4: If m is a bisector of AC, AB = 2x 4, AC = 50, find BC.
BC
m
Recall: AB + BC = AC
AB + BC = AC2x 4 + BC = 502x 4 + 2x 4 = 504x 8 = 504x = 58x = 14.5
2x 4 = BC2(14.5) 4 = BC29 4 = BC25 = BC

Example 5: Given , find JK.
mJK = mLM -2x + 33 = 6x - 23 33 = 8x 23 56 = 8x 7 = x
JK = -2x + 33 JK = -2(7) + 33JK = 19

Now lets switch gears a bit. You should have a good understanding of segments at this point. We will now look at Segment addition. Segment addition problems can sometimes seem confusing, but as long as you remember that the two smaller parts make up the whole, you will be in good shape!

You just need to add the two smaller parts together and set that equal to the total length of the whole segment. Lets look at the equation.
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Notice in the figure below that the two smaller segments, AB and BC, add up to equal AC. You will always use this equation and then just plug in the given information to solve.

Ready for some examples?
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Example 1: If AB = 23 and BC = 47, find AC.
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It may help to draw the figure and label what you know before solving these.

Fill in the parts of the equation you are given and solve. Here we replace AB with 23 and BC with 47. Now, combine like terms and you have 70 = AC. The length of segment AC is 70 units.

Easy enough? Lets do another example.

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It is most helpful to label a drawing before setting up your equation. Here, we see that in Example 2 : AB = 4x+ 5, BC = 15, and AC = 60, find x. We set up the equation 4x + 5 + 15 = 60. Combining like terms on the left we get 4x + 20. Once we subtract 20 form both sides we have 4x = 40. Dividing both sides by 4 to isolate x gives us x = 10 and we have found our solution.

Lets take a look at another example.

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In example 3 we see that AB = 28 and AC = 60. We set up the equation in the form of AB+BC = AC. When we plug in AB and AC we have 28 + BC = 58, but we know we can put an x in for some unknown value, so we replace BC with x and have 28 + x = 58. Subtracting 28 form both sides leaves us with x = 30, or rather BC = 30

Lets try one more example, this one is a bit trickier!

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Here, we see that AB = 2x 8, BC = - x + 24, and AC = 5x 12, and we are asked to find the length of segment AC. We need to set up the equation, but be very, very careful with the signs here! Because the question asks us to find AC, we must first solve for x, then plug that into 5x 12 to find AC.

Our first step is to combine like terms on the left. 2x and x give us a positive 1x. Negative 8 plus 24 gives us 16. Subtract x from each side and we have 16 equals 4x minus 12. Adding 12 to each side leaves 28 = 4x. Finally we divide by 4 and we find that x equals 7.

But dont stop there, we need to find AC. Plug 7 into 5x 12 , using parenthesis, and simplify to find that AC has a length of 23 units.

You can find the length of all 3 parts here by replacing the xs with 7s.

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Midpoints.

We have learned that if A, B, and C are collinear then AB + BC = AC. If AB = BC then B is called the midpoint of AC and we can put in the tic marks to show that AB and BC are congruent. Using this information, we can easily solve segment addition problems that deal with midpoint.
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Example 1: B is the midpoint of AC. If AC = 115 and AB = 5x 10, find x. Because B is a midpoint, we know that AB and BC are congruent. We can use the same equation we have been practicing with which is AB + BC = AC. Because AB is 5x 10 we know that BC = 5x 10. When we plug this into the equation we have 5x 10 + 5x 10 = 115. Combining like terms gives us 10x 20 = 115. Solving for x, we add 20 to each side and then divide by 10 to get x = 13.5.
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Example 2: B is the midpoint of AC. Find x, AB, BC, and AC.

It is given in the figure that AB is 4x + 12 and BC is 5x 3. Since B is the midpoint we know that AB = BC.

We set up the equation as 4x + 12 equals 5x 3 and solve for x.

To keep things positive, I will move the 3 to the left and the 4x to the right. This gives me 15 = x.

We did not use AB + BC = AC in this problem because we are not given any info about AC.
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Now, we can find AB, BC, and AC. To find AB and BC replace x in either AB or BC with 15 (we only need to do this to one because they are congruent when B is the midpoint). We get 4(15) + 12 = AB ; 60 + 12 = AB, 72 = AB. AB = 72 and BC = 72. Using AB + BC = AC we can find AC = 144
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Segment Bisectors. We know what a segment is, but what is a bisector?

Bicycle has TWO wheels that are the same size.Biplane has TWO wings that are the same size.So what do a biplane and a bicycle have in common with a segment bisector?
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A segment Bisector ensures the segment has been divided into TWO parts that are the same size because it crosses through the segment at its midpoint. This means:

Measure of AB equals the measure of BC and AB is congruent to BC and that AB equals one-half of AC as does BC.

Once you know you are dealing with a bisector you can set up the segment addition equation very easily to solve for x or for a missing length, just as we did with midpoint.
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A perpendicular segment Bisector is a segment bisector that runs perpendicular to the segment and passes through the midpoint of the segment.
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Here, we see that m is a bisector of AC, AB = 2x 4, AC = 50, and we need to find BC. Since m is a bisector, we know that AB and BC are congruent. There are several ways to solve this problem, but lets stick to the basic formula we have been using. AB + BC = AC.

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Since AB and BC are congruent, their measures are equal and we can replace BC with 2x 4 in our equation.

This gives us 2x 4 + 2x 4 = 50. Combine like terms on the left, and we have 4x 8 = 50. We move the 8 and have 4x = 58. Dividing out the 4 we are left with x = 14.5.

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Now, to find the length of BC, we replace x in 2x 4 with the 14.5 and find that BC = 25.
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Sometimes, you are just given two segments that are congruent. As you know, this means their measures are the same. In example 5 we have JK is congruent to LM. If JK is -2x + 33 and LM is 6x 23, find JK.

We can write -2x+33 = 6x 23 and solve for x. By moving the 2x and then the 23, we find that x = 7.

Plugging 7 into the x in JK we have JK = 19 units.
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