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9. Properties of Lines & Triangles
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Integrated Math 1

9.06 Perpendicular and angle bisectors

Lesson
Practice
Lesson

Concept summary

When an angle bisector cuts an angle into two congruent angles, we can use the angle bisector theorem and the converse of the angle bisector theorem to solve problems in angles and triangles.

Equidistant

The same distance from two or more objects.

Angle bisector theorem

If a point is on an angle bisector, then it is equidistant from the two legs of the angle.

An angle and a segment. The first endpoint of the segment is on the vertex of the angle, and the second endpoint is in the interior of the angle. The segment divides the angle into two equal angles. From the second vertex of the segment, one segment is drawn to and perpendicular to one of the legs of the angle, and another segment is drawn to and perpendicular to the other leg of the angle.
Converse of angle bisector theorem

If a point is in the interior of an angle and is equidistant from the legs of the angle, then it lies on the angle bisector.

When a perpendicular bisector cuts a line segment at a right angle and into two congruent segments, we can use the perpendicular bisector theorem and the converse of the perpendicular bisector theorem to solve problems in angles and triangles.

Perpendicular bisector theorem

If a point is on the perpendicular bisector of a line segment, then it is equidistant from the end points of the line segment.

A triangle with two congruent legs. A segment is drawn from the apex of the triangle to a point on the base of the triangle, and that is perpendicular to the base. This segment divides the base of the triangle into 2 congruent segments.
Converse of perpendicular bisector theorem

If a point is equidistant from the end points of a line segment, then it is on the perpendicular bisector of that line segment.

Worked examples

Example 1

Construct a perpendicular bisector for the base of the triangle below.

An acute triangle.

Approach

We can construct a perpendicular bisector for a triangle the same way we did for a line segment.

Solution

To construct a perpendicular bisector we can follow the given steps:

  1. Set the compass to a length wider than half of the length of the base of the triangle.

  2. Draw an arc with one vertex at the center.

  3. Draw an arc with the other vertex at the center.

  4. Draw a line through the points where the two arcs intersect.

An acute triangle with the midpoint of the triangle's base shown. A line perpendicular to the base and passing through the midpoint is drawn. Construction lines are drawn.

Reflection

We can also construct a perpendicular bisector using technology. For example, given the triangle shown let's create a perpendicular bisector for the base labeled AB.

A screenshot of the GeoGebra geometry tool showing triangle with base labeled A B and a circle centered at A. Speak to your teacher for more details.

We want to create an arc from point A which has a radius greater than half of the length of the base of the triangle. We then want to create an arc with the same radius centered on point B.

To do so, we can make use of the Circle: Center & Radius tool, to ensure that both arcs have the same radius. We can check the length of \overline{AB} by clicking on the Algebra view to ensure we choose a radius length large enough.

A screenshot of the GeoGebra geometry tool showing the previous image with a new circle centered at B. The points of intersection of the this circle and the previous circle are labeled D and E. Speak to your teacher for more details.

Create a second circle using the Circle: Center & Radius tool with the same radius as the first, this time centered on point B.

Then use the Intersect tool and select the two circles to generate the points D and E.

A screenshot of the GeoGebra geometry tool showing the previous image with a line constructed through points D and E. Speak to your teacher for more details.

Lastly, use the Line tool to create the line through D and E. This is the perpendicular bisector of the base \overline{AB}. Constructions may be hidden using the settings menu in the Algebra view.

Example 2

Find the value of x.

A triangle with leg lengths of 15 and 2 x plus 3. A segment is drawn from the apex of the triangle to a point on the base of the triangle, and that is perpendicular to the base. This segment divides the base of the triangle into 2 congruent segments.

Approach

We can see from the diagram that the vertical line is the perpendicular bisector because it cuts the horizontal line in half and is perpendicular to it.

Solution

Since we have a perpendicular bisector, we can apply the perpendicular bisector theorem which tells us that the lengths with measures of 15 and 2x+3 are equal.

\displaystyle 15\displaystyle =\displaystyle 2x+3 Perpendicular bisector theorem
\displaystyle 12\displaystyle =\displaystyle 2x Subtract 3 from both sides
\displaystyle 6\displaystyle =\displaystyle x Divide both sides by 2

Example 3

Determine whether or not the given diagram is valid. Justify your answer.

A triangle with a segment drawn from the apex of the triangle to a point on the base of the triangle, and that is perpendicular to the base. This segment divides the base of the triangle into 2 congruent segments. The base angles of the triangle have measures of 54 degrees, and 56 degrees.

Solution

The given diagram is not valid.

The angle measures not being equal means that the two non-base sides of the triangle are not congruent, using the converse of the base angles theorem. This means that the vertical line cannot be the perpendicular bisector of the base, using the converse of perpendicular bisector theorem. This contradicts the figure, which shows the vertical line segment being a perpendicular bisector of the base.

Example 4

A mobile phone tower is to be placed equally distant from the three locations shown of the map below.

2
4
6
8
10
12
14
16
18
x \left(\text{mi}\right)
2
4
6
8
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y \left(\text{mi}\right)
a

Using technology, determine the coordinates on the location of the tower correct to one decimal place.

Approach

We can recreate the diagram and then find the required location by doing the following:

  • Construct the perpendicular bisector of \overline{AB} to find all the points equally distant from locations A and B.

  • Construct the perpendicular bisector of \overline{AC} to find all the points equally distant from locations A and C.

  • Identify the intersection point of the two perpendicular bisectors, this is the point equally distant from all three locations.

Solution

A screenshot of the GeoGebra geometry tool showing triangle ABC on a coordinate plane with vertices A (2, 2), B (14, 4) and C (6, 16). Speak to your teacher for more details.

Recreate the image by first adjusting the scale and position of the axes. This can be done by right clicking, selecting settings, and then adjusting the minimum and maximum values for each axis.

Next, to construct the triangle, use the Point tool to add the three points A, B and C. Then use the Segment tool to create the edges between each pair of points.

A screenshot of the GeoGebra geometry tool showing the previous image with the perpendicular bisectors of A B and A C constructed. Speak to your teacher for more details.

Use the Perpendicular Bisector tool and select the points A and B, to create the line of points that are equally distant from the locations A and B.

Use the Perpendicular Bisector tool once more, and select the points A and C, to create the line of points that are equally distant from the locations A and C.

A screenshot of the GeoGebra geometry tool showing the previous image with the intersection point of the two perpendicular bisectors labeled D (7.2, 8.1). Speak to your teacher for more details.

Lastly, use the Intersection tool and select the two perpendicular bisectors to find the location of the tower. To view the coordinates of the point of intersection, from the Algebra view select the settings from the menu for the point of intersection and set the label to include the value.

Thus, the tower should be located at coordinates \left(7.2, 8.1\right) on the plans.

Reflection

If the question had required more decimal places in the answer, we can access the global settings by right clicking and selecting settings, followed by selecting the cog from the icons on the right. Here we can adjust the rounding setting.

A screenshot of the GeoGebra geometry tool showing the global settings menu. Speak to your teacher for more details.
b

Find the distance from the tower to each location. Give your answer correct to one decimal place.

Approach

The tower is located at an equal distance from each location, so select any vertex and find the distance to point D found in part

a
.

Solution

A screenshot of the GeoGebra geometry tool showing the previous image from part b with the distance AD labeled 8. Speak to your teacher for more details.

Use the Distance or Length tool and select points A and D to find the required distance.

The tower is located 8.0 miles from each location.

Outcomes

M1.G.CO.B.3

Use definitions and theorems about lines and angles to solve problems and to justify relationships in geometric figures.

M1.G.CO.C.5

Perform formal geometric constructions with a variety of tools and methods (compass and straightedge, string, reflective devices, paper folding, dynamic geometric software, etc.).

M1.G.CO.C.6

Use geometric constructions to solve geometric problems in context, by hand and using technology.*

M1.MP1

Make sense of problems and persevere in solving them.

M1.MP2

Reason abstractly and quantitatively.

M1.MP3

Construct viable arguments and critique the reasoning of others.

M1.MP4

Model with mathematics.

M1.MP5

Use appropriate tools strategically.

M1.MP6

Attend to precision.

M1.MP7

Look for and make use of structure.

M1.MP8

Look for and express regularity in repeated reasoning.

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