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1.4: Midpoints and Bisectors

Difficulty Level: At Grade Created by: CK-12
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Learning Objectives

  • Identify the midpoint of line segments.
  • Identify the bisector of a line segment.
  • Understand and the Angle Bisector Postulate.

Review Queue

Answer the following questions.

  1. \begin{align*}m \angle ROT = 165^\circ\end{align*}, find \begin{align*}m \angle POT\end{align*}
  2. Find \begin{align*}x\end{align*}.
  3. Use the Angle Addition Postulate to write an equation for the angles in #1.

Know What? The building to the right is the TransamericaBuilding in San Francisco. This building was completed in 1972 and, at that time was one of the tallest buildings in the world. It is a pyramid with two “wings” on either side, to accommodate elevators. Because San Francisco has problems with earthquakes, there are regulations on how a building can be designed. In order to make this building as tall as it is and still abide by the building codes, the designer used this pyramid shape.

It is very important in designing buildings that the angles and parts of the building are equal. What components of this building look equal? Analyze angles, windows, and the sides of the building.


You could argue that another word for equal is congruent. However, the two differ slightly.

Congruent: When two geometric figures have the same shape and size.

We label congruence with a \begin{align*}\cong\end{align*} sign. Notice the \begin{align*}\sim\end{align*} above the = sign. \begin{align*}\overline{AB} \cong \overline{BA}\end{align*} means that \begin{align*}\overline{AB}\end{align*} is congruent to \begin{align*}\overline{BA}\end{align*}. If we know two segments or angles are congruent, then their measures are also equal. If two segments or angles have the same measure, then, they are also congruent.

Equal Congruent
\begin{align*}=\end{align*} \begin{align*}\cong\end{align*}
used with measurement used with figures
\begin{align*}m \overline{AB} = AB = 5 \ cm\end{align*} \begin{align*}\overline{AB} \cong \overline{BA}\end{align*}
\begin{align*}m \angle ABC = 60^\circ\end{align*} \begin{align*}\angle ABC \cong \angle CBA\end{align*}


Midpoint: A point on a line segment that divides it into two congruent segments.

Because \begin{align*}AB = BC, \ B\end{align*} is the midpoint of \begin{align*}\overline{AC}\end{align*}.

Midpoint Postulate: Any line segment will have exactly one midpoint.

This might seem self-explanatory. However, be careful, this postulate is referring to the midpoint, not the lines that pass through the midpoint, which is infinitely many.

Example 1: Is \begin{align*}M\end{align*} a midpoint of \begin{align*}\overline{AB}\end{align*}?

Solution: No, it is not because \begin{align*}MB = 16\end{align*} and \begin{align*}AM = 34 - 16 = 18\end{align*}.

Midpoint Formula

When points are plotted in the coordinate plane, you can use slope to find the midpoint between then. We will generate a formula here.

Here are two points, (-5, 6) and (3, 4). Draw a line between the two points and determine the vertical distance and the horizontal distance.

So, it follows that the midpoint is down and over half of each distance. The midpoint would then be down 2 (or -2) from (-5, 6) and over positively 4. If we do that we find that the midpoint is (-1, 4).

Let’s create a formula from this. If the two endpoints are (-5, 6) and (3, 4), then the midpoint is (-1, 4). -1 is halfway between -5 and 3 and 4 is halfway between 6 and 2. Therefore, the formula for the midpoint is the average of the \begin{align*}x-\end{align*}values and the average of the \begin{align*}y-\end{align*}values.

Midpoint Formula: For two points, \begin{align*}(x_1, \ y_1)\end{align*} and \begin{align*}(x_2, \ y_2)\end{align*}, the midpoint is \begin{align*}\left ( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right )\end{align*}

Example 2: Find the midpoint between (9, -2) and (-5, 14).

Solution: Plug the points into the formula.

\begin{align*} \left ( \frac{9 + (-5)}{2}, \ \frac{-2 + 14}{2} \right ) = \left ( \frac{4}{2}, \ \frac{12}{2} \right ) = (2, 6)\end{align*}

Example 3: If \begin{align*}M(3, -1)\end{align*} is the midpoint of \begin{align*}\overline{AB}\end{align*} and \begin{align*}B(7, -6)\end{align*}, find \begin{align*}A\end{align*}.

Solution: Plug what you know into the midpoint formula.

\begin{align*}\left ( \frac{7 + x_A}{2}, \frac{-6 + y_A}{2} \right ) &= (3, -1)\\ \frac{7 + x_A}{2} &= 3 \ \text{and} \ \frac{-6 + y_A}{2} = -1 \qquad \ A \ \text{is} \ (-1, 4).\\ 7 + x_A &= 6 \ \text{and} \ -6 + y_A = -2\\ x_A &= -1 \ \text{and} \ y_A = 4\end{align*}

Another way to find the other endpoint is to find the difference between \begin{align*}M\end{align*} and \begin{align*}B\end{align*} and then duplicate it on the other side of \begin{align*}M\end{align*}.

\begin{align*}x-\end{align*}values: \begin{align*}7 - 3 = 4\end{align*}, so 4 on the other side of 3 is \begin{align*}3 - 4 = -1\end{align*}

\begin{align*}y-\end{align*}values: \begin{align*}-6 -(-1) = -5\end{align*}, so -5 on the other side of -1 is \begin{align*}-1 - (-5) = 4\end{align*}

\begin{align*}A\end{align*} is still (-1, 4). You may use either method.

Segment Bisectors

Segment Bisector: A line, segment, or ray that passes through a midpoint of another segment.

A bisector cuts a line segment into two congruent parts.

Example 4: Use a ruler to draw a bisector of the segment below.

Solution: The first step in identifying a bisector is finding the midpoint. Measure the line segment and it is 4 cm long. To find the midpoint, divide 4 by 2.

So, the midpoint will be 2 cm from either endpoint, or halfway between. Measure 2 cm from one endpoint and draw the midpoint.

To finish, draw a line that passes through the midpoint. It doesn’t matter how the line intersects \begin{align*}\overline{XY}\end{align*}, as long as it passes through \begin{align*}Z\end{align*}.

A specific type of segment bisector is called a perpendicular bisector.

Perpendicular Bisector: A line, ray or segment that passes through the midpoint of another segment and intersects the segment at a right angle.

\begin{align*}\overleftrightarrow{DE}\end{align*} is the perpendicular bisector of \begin{align*}\overline{AC}\end{align*}, so \begin{align*}\overline{AB} \cong \overline{BC}\end{align*} and \begin{align*}\overline{AC} \bot \overleftrightarrow{DE}\end{align*}.

Perpendicular Bisector Postulate: For every line segment, there is one perpendicular bisector that passes through the midpoint.

There are infinitely many bisectors, but only one perpendicular bisector for any segment.

Example 5: Which line is the perpendicular bisector of \begin{align*}\overline{MN}\end{align*}?

Solution: The perpendicular bisector must bisect \begin{align*}\overline{MN}\end{align*} and be perpendicular to it. Only \begin{align*}\overleftrightarrow{OQ}\end{align*} satisfies both requirements. \begin{align*}\overleftrightarrow{SR}\end{align*} is just a bisector.

Example 6: Algebra Connection Find \begin{align*}x\end{align*} and \begin{align*}y\end{align*}.

Solution: The line shown is the perpendicular bisector. So, \begin{align*}3x - 6 = 21, \ 3x = 27, \ x = 9\end{align*}.

And, \begin{align*}(4y - 2)^\circ = 90^\circ, 4y^\circ = 92^\circ, y = 23^\circ\end{align*}.

Investigation 1-3: Constructing a Perpendicular Bisector

  1. Draw a line that is at least 6 cm long, about halfway down your page.
  2. Place the pointer of the compass at an endpoint. Open the compass to be greater than half of the segment. Make arc marks above and below the segment. Repeat on the other endpoint. Make sure the arc marks intersect.
  3. Use your straight edge to draw a line connecting the arc intersections.

This constructed line bisects the line you drew in #1 and intersects it at \begin{align*}90^\circ\end{align*}. So, this construction also works to create a right angle. To see an animation of this investigation, go to http://www.mathsisfun.com/geometry/construct-linebisect.html.

Congruent Angles

Example 7: Algebra Connection What is the measure of each angle?

Solution: From the picture, we see that the angles are congruent, so the given measures are equal.

\begin{align*}(5x + 7)^\circ & = (3x + 23)^\circ\\ 2x^\circ & = 16^\circ\\ x & = 8^\circ\end{align*}

To find the measure of \begin{align*}\angle ABC\end{align*}, plug in \begin{align*}x = 8^\circ\end{align*} to \begin{align*}(5x + 7)^\circ\end{align*}.

\begin{align*}& (5(8) + 7)^\circ\\ & (40 + 7)^\circ\\ & 47^\circ\end{align*}

Because \begin{align*}m \angle ABC = m \angle XYZ, \ m \angle XYZ = 47^\circ\end{align*} too.

Angle Bisectors

Angle Bisector: A ray that divides an angle into two congruent angles, each having a measure exactly half of the original angle.

\begin{align*}\overline{BD}\end{align*} is the angle bisector of \begin{align*}\angle ABC\end{align*}

\begin{align*}\angle ABD & \cong \angle DBC\\ m \angle ABD & = \frac{1}{2} m \angle ABC\end{align*}

Angle Bisector Postulate: Every angle has exactly one angle bisector.

Example 8: Let’s take a look at Review Queue #1 again. Is \begin{align*}\overline{OP}\end{align*} the angle bisector of \begin{align*} \angle SOT\end{align*}? Recall, that \begin{align*}m \angle ROT = 165^\circ\end{align*}, what is \begin{align*}m \angle SOP\end{align*} and \begin{align*}m \angle POT\end{align*}?

Solution: Yes, \begin{align*}\overline{OP}\end{align*} is the angle bisector of \begin{align*}\angle SOT\end{align*} according to the markings in the picture. If \begin{align*}m \angle ROT = 165^\circ\end{align*} and \begin{align*}m \angle ROS = 57^\circ\end{align*}, then \begin{align*}m \angle SOT = 165^\circ - 57^\circ = 108^\circ\end{align*}. The \begin{align*}m \angle SOP\end{align*} and \begin{align*}m \angle POT\end{align*} are each half of \begin{align*}108^\circ\end{align*} or \begin{align*}54^\circ\end{align*}.

Investigation 1-4: Constructing an Angle Bisector

  1. Draw an angle on your paper. Make sure one side is horizontal.
  2. Place the pointer on the vertex. Draw an arc that intersects both sides.
  3. Move the pointer to the arc intersection with the horizontal side. Make a second arc mark on the interior of the angle. Repeat on the other side. Make sure they intersect.
  4. Connect the arc intersections from #3 with the vertex of the angle.

To see an animation of this construction, view http://www.mathsisfun.com/geometry/construct-anglebisect.html.

Know What? Revisited The image to the right is an outline of the Transamerica Building from earlier in the lesson. From this outline, we can see the following parts are congruent:

\begin{align*}&\overline{TR} \cong \overline{TC} &&&& \angle TCR \cong \angle TRC\\ &\overline{RS} \cong \overline{CM} &&&& \angle CIE \cong \angle RAN\\ &\overline{CI} \cong \overline{RA} && \text{and} && \angle TMS \cong \angle TSM\\ &\overline{AN} \cong \overline{IE} &&&& \angle IEC \cong \angle ANR\\ &\overline{TS} \cong \overline{TM} &&&& \angle TCI \cong \angle TRA\end{align*}

As well at these components, there are certain windows that are congruent and all four triangular sides of the building are congruent to each other.

Review Questions

  1. Copy the figure below and label it with the following information:

\begin{align*}\angle A \cong \angle C\\ \angle B \cong \angle D\\ \overline{AB} \cong \overline{CD}\\ \overline{AD} \cong \overline{BC}\end{align*}

For 2-9, find the lengths, given: \begin{align*}H\end{align*} is the midpoint of \begin{align*}\overline{AE}\end{align*} and \begin{align*}\overline{DG}, \ B\end{align*} is the midpoint of \begin{align*}\overline{AC}, \ \overline{GD}\end{align*} is the perpendicular bisector of \begin{align*}\overline{FA}\end{align*} and \begin{align*}\overline{EC}, \ \overline{AC} \cong \overline{FE},\end{align*} and \begin{align*}\overline{FA} \cong \overline{EC}\end{align*}.

  1. \begin{align*}AB\end{align*}
  2. \begin{align*}GA\end{align*}
  3. \begin{align*}ED\end{align*}
  4. \begin{align*}HE\end{align*}
  5. \begin{align*}m \angle HDC\end{align*}
  6. \begin{align*}FA\end{align*}
  7. \begin{align*}GD\end{align*}
  8. \begin{align*}m \angle FED\end{align*}
  9. How many copies of triangle \begin{align*}AHB\end{align*} can fit inside rectangle \begin{align*}FECA\end{align*} without overlapping?

For 11-18, use the following picture to answer the questions.

  1. What is the angle bisector of \begin{align*} \angle TPR\end{align*}?
  2. \begin{align*}P\end{align*} is the midpoint of what two segments?
  3. What is \begin{align*}m \angle QPR\end{align*}?
  4. What is \begin{align*}m \angle TPS\end{align*}?
  5. How does \begin{align*} \overline{VS}\end{align*} relate to \begin{align*} \overline{QT}\end{align*}?
  6. How does \begin{align*} \overline{QT}\end{align*} relate to \begin{align*} \overline{VS}\end{align*}?
  7. Is \begin{align*} \overline{PU}\end{align*} a bisector? If so, of what?
  8. What is \begin{align*}m \angle QPV\end{align*}?

Algebra Connection For 19-24, use algebra to determine the value of variable(s) in each problem.

  1. Construction Using your protractor, draw an angle that is \begin{align*}110^\circ\end{align*}. Then, use your compass to construct the angle bisector. What is the measure of each angle?
  2. Construction Using your protractor, draw an angle that is \begin{align*}75^\circ\end{align*}. Then, use your compass to construct the angle bisector. What is the measure of each angle?
  3. Construction Using your ruler, draw a line segment that is 7 cm long. Then use your compass to construct the perpendicular bisector, What is the measure of each segment?
  4. Construction Using your ruler, draw a line segment that is 4 in long. Then use your compass to construct the perpendicular bisector, What is the measure of each segment?
  5. Construction Draw a straight angle \begin{align*}(180^\circ)\end{align*}. Then, use your compass to construct the angle bisector. What kind of angle did you just construct?

For questions 30-33, find the midpoint between each pair of points.

  1. (-2, -3) and (8, -7)
  2. (9, -1) and (-6, -11)
  3. (-4, 10) and (14, 0)
  4. (0, -5) and (-9, 9)

Given the midpoint \begin{align*}(M)\end{align*} and either endpoint of \begin{align*}\overline{AB}\end{align*}, find the other endpoint.

  1. \begin{align*}A(-1, 2)\end{align*} and \begin{align*}M(3, 6)\end{align*}
  2. \begin{align*}B(-10, -7)\end{align*} and \begin{align*}M(-2, 1)\end{align*}
  3. Error Analysis Erica is looking at a geometric figure and trying to determine which parts are congruent. She wrote \begin{align*}\overline{AB} = \overline{CD}\end{align*}. Is this correct? Why or why not?
  4. Challenge Use the Midpoint Formula to solve for the \begin{align*}x-\end{align*}value of the midpoint and the \begin{align*}y-\end{align*}value of the midpoint. Then, use this formula to solve #34. Do you get the same answer?
  5. Construction Challenge Use construction tools and the constructions you have learned in this section to construct a \begin{align*}45^\circ\end{align*} angle.
  6. Construction Challenge Use construction tools and the constructions you have learned in this section to construct two 2 in segments that bisect each other. Now connect all four endpoints with segments. What figure have you constructed?
  7. Describe an example of how the concept of midpoint (or the midpoint formula) could be used in the real world.

Review Queue Answers

  1. See Example 6
  2. \begin{align*}2x - 5 = 33\!\\ {\;} \quad \ 2x = 38\!\\ {\;} \quad \ \ x = 19\end{align*}
  3. \begin{align*}m \angle ROT = m \angle ROS + m \angle SOP + m \angle POT\end{align*}

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