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# 1.4: Midpoints and Bisectors

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

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

## Review Queue

1. mSOP=38\begin{align*}m\angle SOP = 38^\circ\end{align*}, find mPOT\begin{align*}m\angle POT\end{align*} and mROT\begin{align*}m\angle ROT\end{align*}.
2. Find the slope between the two numbers.
1. (-4, 1) and (-1, 7)
2. (5, -6) and (-3, -4)
3. Find the average of these numbers: 23, 30, 18, 27, and 32.

Know What? The building to the right is the Transamerica Building in San Francisco. This building was completed in 1972 and, at that time was one of the tallest buildings in the world. 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.

## Congruence

You could argue that another word for equal is congruent. But, the two are a little different.

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

Label It Say It
AB¯¯¯¯¯¯¯¯BA¯¯¯¯¯¯¯¯\begin{align*}\overline{AB} \cong \overline{BA}\end{align*} AB\begin{align*}AB\end{align*} is congruent to BA\begin{align*}BA\end{align*}
Equal Congruent
=\begin{align*}=\end{align*} \begin{align*}\cong\end{align*}
used with measurement used with figures
mAB¯¯¯¯¯¯¯¯=AB=5cm\begin{align*}m\overline{AB} = AB = 5 cm\end{align*} AB¯¯¯¯¯¯¯¯BA¯¯¯¯¯¯¯¯\begin{align*}\overline{AB} \cong \overline{BA}\end{align*}
mABC=60\begin{align*}m\angle ABC = 60^\circ\end{align*} ABCCBA\begin{align*}\angle ABC \cong \angle CBA\end{align*}

If two segments or angles are congruent, then they are also equal.

## Midpoints

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

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

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

This postulate is referring to the midpoint, not the lines that pass through the midpoint.

There are infinitely many lines that pass through the midpoint.

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

Solution: No, it is not MB=16\begin{align*}MB = 16\end{align*} and AM=3416=18\begin{align*}AM = 34 - 16 = 18\end{align*}. AM\begin{align*}AM\end{align*} must equal MB\begin{align*}MB\end{align*} in order for M\begin{align*}M\end{align*} to be the midpoint of AB¯¯¯¯¯¯¯¯\begin{align*}\overline{AB}\end{align*}.

## Midpoint Formula

When points are plotted in the coordinate plane, we can use a formula to find the midpoint between them. Here are two points, (-5, 6) and (3, 4).

It follows that the midpoint should be halfway between the points on the line. Just by looking, it seems like the midpoint is (-1, 4).

Midpoint Formula: For two points, (x1,y1)\begin{align*}(x_1, y_1)\end{align*} and x2,y2\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*}.

Let’s use the formula to make sure (-1, 4) is the midpoint between (-5, 6) and (3, 2).

\begin{align*}\left ( \frac{-5+3}{2}, \ \frac{6+2}{2} \right ) = \left (\frac{-2}{2}, \frac{8}{2} \right ) = (-1, 4)\end{align*}

Always use this formula to determine the midpoint.

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 in 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\\ &7 + x_A = 6 \ \text{and} \ -6 + y_A = -2\\ & x_A = -1 \ \text{and} \ y_A = 4\\ &\text{So}, \ A \ \text{is} \ (-1, 4).\end{align*}

## Segment Bisectors

Segment Bisector: A bisector cuts a line segment into two congruent parts and passes through the midpoint.

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

Solution: First, find the midpoint. Measure the line segment. It is 4 cm long. To find the midpoint, divide 4 cm by 2 because we want 2 equal pieces. Measure 2 cm from one endpoint and draw the midpoint.

To finish, draw a line that 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*}&\overline{AB} \cong \overline{BC}\\ &\overline{AC} \perp \overleftrightarrow{DE}\end{align*}

Perpendicular Bisector Postulate: For every line segment, there is one perpendicular bisector.

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*} fits this description. \begin{align*}\overleftrightarrow{SR}\end{align*} is a bisector, but is not perpendicular.

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

Solution: The line shown is the perpendicular bisector.

\begin{align*}\text{So}, \ 3x - 6 &= 21 && \text{And}, \ (4y - 2)^\circ = 90^\circ\\ 3x &= 27 && \qquad \qquad \ \ \quad 4y = 92^\circ\\ x &= 9 && \qquad \qquad \quad \ \ \ y = 23^\circ\end{align*}

Investigation 1-4: Constructing a Perpendicular Bisector

1. Draw a line that is 6 cm long, 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 straightedge 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*}. 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 equal.

Set the angles equal to each other and solve.

\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 \rightarrow (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*}?

Solution: Yes, \begin{align*}\overline{OP}\end{align*} is the angle bisector of \begin{align*}\angle SOT\end{align*} from the markings in the picture.

Investigation 1-5: 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} \qquad \qquad \quad \angle TCR \cong \angle TRC\\ &\overline{RS} \cong \overline{CM} \qquad \qquad \quad \angle CIE \cong \angle RAN\\ &\overline{CI} \cong \overline{RA} \qquad \text{and} \quad \ \angle TMS \cong \angle TSM\\ &\overline{AN} \cong \overline{IE} \qquad \qquad \quad \angle IEC \cong \angle ANR\\ &\overline{TS} \cong \overline{TM} \qquad \qquad \quad \angle TCI \cong \angle TRA\end{align*}

All the four triangular sides of the building are congruent to each other as well.

## Review Questions

• Questions 1-18 are similar to Examples 1, 4, 5 and 8.
• Questions 19-22 are similar to Examples 6 and 7.
• Question 23 is similar to Investigation 1-5.
• Question 24 is similar to Investigation 1-4.
• Questions 25-28 are similar to Example 2.
• Question 29 and 30 are similar to Example 3.
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*}

\begin{align*}H\end{align*} is the midpoint of \begin{align*}\overline{AE}\end{align*} and \begin{align*}\overline{DG}\end{align*}, \begin{align*}B\end{align*} is the midpoint of \begin{align*}\overline{AC}\end{align*}, \begin{align*}\overline{GD}\end{align*} is the perpendicular bisector of \begin{align*}\overline{FA}\end{align*} and \begin{align*}\overline{EC}\end{align*} \begin{align*}\overline{AC} \cong \overline{FE}\end{align*} and \begin{align*}\overline{FA} \cong \overline{EC}\end{align*}

Find:

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. What is \begin{align*}m\angle QPV\end{align*}?
8. Is \begin{align*}\overline{PU}\end{align*} a bisector? If so, of what?

Algebra Connection For 19-22, use algebra to determine the value of variable 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 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?

For questions 25-28, 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*}

1. \begin{align*}m\angle POT = 38^\circ, \ m\angle ROT = 57^\circ + 38^\circ + 38^\circ = 133^\circ\end{align*}
1. \begin{align*}\frac{7-1}{-1+4} = \frac{6}{3} = 2\end{align*}
2. \begin{align*}\frac{-4+6}{-3-5} = \frac{2}{-8} = - \frac{1}{4}\end{align*}
2. \begin{align*}\frac{23+30+18+27+32}{5} = \frac{130}{5} = 26\end{align*}

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