Construct the midpoint or perpendicular bisector of a segment

Key Notes :

A perpendicular bisector splits a segment into two congruent segments and is perpendicular to that segment. So, the perpendicular bisector of a segment passes through the midpoint of the segment at a 90° angle.

For example, CD⟂AB at point M and AM≅BM. So, CD is the perpendicular bisector of AB.

Line segment AB and line CD are shown. Line CD intersects segment AB at point M, creating 90 degree angles. Segment AM is congruent to segment BM.

Perpendicular Bisector Theorem

The Perpendicular Bisector Theorem states that if a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.

For example, the diagram below shows that point Q is on the perpendicular bisector of PR.

Triangle PQR is shown. Segment PR is intersected by line QS at point S, creating 90 degree angles. Segment PS is congruent to segment RS.

So, point Q is equidistant from points P and R. That means that PQ≅QR.

Triangle PQR is shown. Segment PR is intersected by line QS at point S, creating 90 degree angles. Segment PS is congruent to segment RS. Segment PQ is congruent to segment QR.

Using the Perpendicular Bisector Theorem

You can use the Perpendicular Bisector Theorem to solve for missing side lengths in triangles.

Let’s try it! Find the length of KL.

Triangle JKL is shown. Segment JL is intersected by line KM at point M, creating 90 degree angles. Segment JM is congruent to segment LM. Segment JK is labeled 6.1.

Since ∠JMK is a right angle and JM≅LM, KM is the perpendicular bisector of JL. This means that point K is on the perpendicular bisector. Therefore, JK≅KL by the Perpendicular Bisector Theorem.

So, JK=KL=6.1.

Another example

Let’s try another example! Find the value of t.

Triangle WXY is shown. Segment WY is intersected by line XZ at point Z, creating 90 degree angles. Segment WZ is congruent to segment YZ. Segment XW is labeled 2t plus 4. Segment XY is labeled 3t minus 7.

Since ∠WZX is a right angle and WZ≅YZ, XZ is the perpendicular bisector of WY. Therefore, WX≅XY by the Perpendicular Bisector Theorem.

So, you can set WX equal to XY and solve for t.

WX=XY

2t+4=3t–7 Plug in WX = 2t + 4 and XY = 3t – 7.

2t+11=3t Add 7 to both sides.

11=t Subtract 2t from both sides.

So, t=11.

Converse of the Perpendicular Bisector Theorem

The Converse of the Perpendicular Bisector Theorem states that if a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.

For example, the diagram below shows that WX≅XY, meaning that point X is equidistant from the endpoints of WY.

Triangle WXY is shown. Segment WX is congruent to segment XY.

So, point X is on the perpendicular bisector of WY.

Triangle WXY is shown. Segment WX is congruent to segment XY. Segment WY is intersected at a 90 degree angle by a line that includes point X. The line splits segment WY into two congruent segments.

Using the Converse of the Perpendicular Bisector Theorem

You can use the Converse of the Perpendicular Bisector Theorem to solve for missing side lengths in triangles.

Let’s try it! Find the length of DG.

Triangle DEF is shown. Segment DE and segment EF are both labeled 52. Segment DF is intersected by line GE at point G, creating 90 degree angles. Segment DG is labeled 7b minus 15. Segment FG is labeled 4b plus 9.

Since DE=EF=52, point E is equidistant from the endpoints of DF. Therefore, point E is on the perpendicular bisector of DF by the Converse of the Perpendicular Bisector Theorem.

You also know that ∠EGF is a right angle. So, EG is the perpendicular bisector of DF. That means that DG≅FG. You can set DG equal to FG and solve for b.

DG=FG

7b–15=4b+9 Plug in DG = 7b – 15 and FG = 4b + 9

7b=4b+24 Add 15 to both sides.

3b=24 Subtract 4b from both sides.

b=8 Divide both sides by 3.

Now, substitute b=8 into 7b–15 to find the length of DG.

7b–15

7(8)–15

56–15

41

So, DG=41.

Constructing perpendicular bisectors

You can construct a perpendicular bisector using a straightedge and a compass.

Start with the segment you want to bisect.

Line segment AB is shown.

Now, set your compass to a setting greater than 1/2 AB and place the point of your compass at point A. Draw arcs above and below AB.

Line segment AB is shown. There are arcs above and below segment AB that were created using a compass with its point placed on point A and a setting greater than half the length of segment AB.

Next, using the same compass setting, place the point of your compass at point B. Draw arcs above and below AB.

Line segment AB is shown. There are arcs above and below segment AB that were created using a compass with its point placed on point A and a setting greater than half the length of segment AB. There are also arcs above and below segment AB that were created using the same compass setting with the compass point placed on point B.

Then, mark the points where the arcs intersect. Call them points C and D.

Line segment AB is shown. There are arcs above and below segment AB that were created using a compass with its point placed on point A and a setting greater than half the length of segment AB. There are also arcs above and below segment AB that were created using the same compass setting with the compass point placed on point B. Point C is the intersection point of the arcs above segment AB. Point D is the intersection point of the arcs below segment AB.

Last, draw a line that connects points C and D using your straightedge.

Line segment AB is shown. There are arcs above and below segment AB that were created using a compass with its point placed on point A and a setting greater than half the length of segment AB. There are also arcs above and below segment AB that were created using the same compass setting with the compass point placed on point B. Point C is the intersection point of the arcs above segment AB. Point D is the intersection point of the arcs below segment AB. There is a line that goes through points C and D.

CD is the perpendicular bisector of AB!

Why does this construction work?

Recall that the Converse of the Perpendicular Bisector Theorem says that if a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment.

In the steps above, if you drew the entire circle for each pair of arcs, the circles would be centered at point A and point B.

Line segment AB is shown. There are two congruent circles, one centered at point A and the other centered at point B. Point C is the intersection point of the circles above segment AB. Point D is the intersection point of the circles below segment AB. There is a line that goes through points C and D.

⨀A and ⨀B were created using the same compass setting, so they are congruent. Since the circles are congruent and point C is on both of them, AC=BC.

Line segment AB is shown. There are two congruent circles, one centered at point A and the other centered at point B. Point C is the intersection point of the circles above segment AB. Point D is the intersection point of the circles above segment AB. There is a line that goes through points C and D. Segment AC is congruent to segment BC.

So, by the Converse of the Perpendicular Bisector Theorem, point C is on the perpendicular bisector of AB.

Similarly, since ⨀A and ⨀B are congruent and point D is on both of them, AD=BD.

Line segment AB is shown. There are two congruent circles, one centered at point A and the other centered at point B. Point C is the intersection point of the circles above segment AB. Point D is the intersection point of the circles above segment AB. There is a line that goes through points C and D. Segment AD is congruent to segment BD.

So, by the Converse of the Perpendicular Bisector Theorem, point D is on the perpendicular bisector of AB.

Since both points C and D are on the perpendicular bisector of AB, CD is the perpendicular bisector of AB.

Line segment AB is shown. There are two congruent circles, one centered at point A and the other centered at point B. Point C is the intersection point of the circles above segment AB. Point D is the intersection point of the circles above segment AB. There is a line that goes through points C and D. Line CD passes through the midpoint of segment AB at a 90 degree angle.

Learn with an example

Mark the midpoint of AB ?

Part of the construction was done for you. Here are the steps to create this part of the construction.

Start with the objects in the diagram below.

  • Draw a circle with radius AB centred at A.
  • Draw a circle with radius AB centred at B.
  • Mark the points where ⨀A and ⨀B intersect. Call them C and D.

Since B and C are both on ⨀A, AB=AC. Since A and C are both on ⨀B, AB=BC. So, AB=AC=BC.

Similarly, AB=AD=BD.

So, C and D are equidistant from A and B.

  • Draw the line through C and D.

Recall that the set of points equidistant from A and B form the perpendicular bisector of AB. Since C and D are equidistant from A and B, they lie on the perpendicular bisector. So , CD is the perpendicular bisector of AB.

Complete the construction.

To complete the construction of the midpoint o f AB , carry out the following step:

  • Mark the point where CD and AB intersect. Call it E.

Since CD is the perpendicular bisector of AB, the intersection E is the midpoint of AB.

Mark the midpoint of AB?

Part of the construction was done for you. Here are the steps to create this part of the construction.

Start with the objects in the diagram below.

  • Draw a circle with radius AB centred at A.
  • Draw a circle with radius AB centred at B.
  • Mark the points where ⨀A and ⨀B intersect. Call them C and D.

Since B and C are both on ⨀A, AB=AC. Since A and C are both on ⨀B, AB=BC. So, AB=AC=BC.

Similarly, AB=AD=BD.

So, C and D are equidistant from A and B.

  • Draw the line through C and D.

Recall that the set of points equidistant from A and B form the perpendicular bisector of AB . Since C and D are equidistant from A and B, they lie on the perpendicular bisector. So , CD is the perpendicular bisector of AB.

Complete the construction.

To complete the construction of the midpoint of AB , carry out the following step:

  • Mark the point where CD and AB intersect. Call it E.

Since CD is the perpendicular bisector of AB , the intersection E is the midpoint of AB.

let’s practice!