7. Triangles

7.02 Isosceles and equilateral triangles

7.03 Angle and perpendicular bisectors

Lesson

Practice

7.04 Medians and altitudes

7.05 Midsegments

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United States of America

Grades 9-12

7.03 Angle and perpendicular bisectors

Lesson

Practice

Introduction

We first constructed angle bisectors in lesson 1.04 Angles and constructions and perpendicular bisectors in lesson 2.03 Perpendicular lines. We will construct angle bisectors and perpendicular bisectors in triangles in this lesson to solve problems and draw constructions.

Perpendicular bisectors

Equidistant

The same distance from two or more objects

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.

Segment M N with a point P on M N. Another segment O P is drawn perpendicular to M N. Points Q, R, S and T are on O P. For each point, dashed segment are drawn from the point to M and N.

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

Exploration

Draw three different triangles and construct the perpendicular bisectors of each side. Make sure each triangle is a different type: acute, obtuse, and right.

What do you notice about the perpendicular bisectors of the first triangle?
Does what you noticed with the first triangle hold true for the other two?

A point of concurrency is a point where three or more lines coincide. The perpendicular bisectors of a triangle intersect at the point of concurrency, called the circumcenter. The circumcenter is also the center of the circle that circumscribes the triangle.

Circumscribed circle

The smallest circle that includes a plane figure. If the figure is a polygon, then the circle must contain all of the vertices of the polygon.

A triangle inside a circle. Each vertex of the triangle lies on the circle.

Circumcenter

The point of concurrency of the perpendicular bisectors of the triangle. It is called the circumcenter because it is the center of the circumscribed circle of the triangle.

A triangle where each side has a line passing through it. Each line splits the side it passes through exactly in half and intersects the side at a 90 degree angle. All 3 lines intersect at a single point

Circumcenter theorem

The perpendicular bisectors of a triangle intersect at a point that is equidistant from the vertices of the triangle.

A triangle inside a circle. Each vertex of the triangle lies on the circle. Each side of the triangle has a line passing through it. Each line splits the side it passes through exactly in half and perpendicular to a side. All 3 lines intersect at a single point inside the triangle.

When the circumcenter is located inside the triangle, the triangle is acute.

When the circumcenter is located outside of the triangle, the triangle is obtuse.

When the circumcenter is located on the triangle, the triangle is right. Note that the circumcenter is the midpoint of the hypotenuse.

Examples

Example 1

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.

Worked Solution

Create a strategy

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.

Apply the idea

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 2

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

x \left(\text{mi}\right)

y \left(\text{mi}\right)

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

Worked Solution

Create a strategy

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.

Apply the idea

A screenshot of the GeoGebra geometry tool showing triangle A B C. 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 how to construct perpendicular bisectors of triangle A B C. 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 how to find point of intersection of the perpendicular bisectors of triangle A B C. 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.

Reflect and check

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 how to adjust the rounding setting. Speak to your teacher for more details.

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

Worked Solution

Create a strategy

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).

Apply the idea

A screenshot of the GeoGebra geometry tool showing how to find the distance between point D and vertex A. 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.

Example 3

Construct a circumscribed circle for the triangle below:

Worked Solution

Create a strategy

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

Apply the idea

Set the compass to a length wider than half of the length of the base of the triangle, draw an arc with one vertex at A. Then, using the same compass setting, draw an arc with the other vertex at C. Draw a line through the points where the two arcs intersect.
Repeat Step 1 for another segment on the triangle.
Repeat Step 1 for the final segment on the triangle.
Set the compass to the length of the circumcenter and one of the vertices on the triangle, then draw a circle.
Erase all construction marks.

Idea summary

The circumcenter of a triangle the point of concurrency of the perpendicular bisectors of the triangle. It is also the center of the circumscribed circle of the triangle.

The circumcenter will lie inside of an acute triangle, outside of an obtuse triangle, and on the midpoint of the hypotenuse of a right triangle. The circumcenter will be equidistant from each vertex of a triangle.

To construct a circumscribed circle of a triangle, we can construct perpendicular bisectors on each segment. Then, we can draw a circle using their point of concurrency and a vertex on the triangle as the radius.

Angle bisectors

Recall that an angle bisector is a line, segment or ray that divides an angle into two congruent angles. 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.

Angle bisector theorem

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

While we can construct a variety of angle bisectors and verify these theorems through experimentation, we don't yet have the knowledge to write a formal proof. We will formally prove the angle bisector theorem and its converse in a later lesson.

Exploration

Draw three different triangles and construct the angle bisectors of each side. Make sure each triangle is a different type: acute, obtuse, and right.

What do you notice about the angle bisectors of the first triangle?
Does what you noticed with the first triangle hold true for the other two?

The angle bisectors of a triangle intersect at a point of concurrency, called the incenter. The incenter is also the center of the inscribed circle of a triangle.

Inscribed circle

The largest possible circle that can be contained in a plane figure. If the plane figure is a polygon, then the circle must be tangent to all of the sides of the polygon.

A circle inside of a triangle. Each side of the triangle intersects the circle at exactly one point.

Incenter

The point of concurrency of the angle bisectors of a triangle. It is called the incenter because it is the center of the inscribed circle of the triangle.

A triangle with a dashed line segment coming from each angle and intersecting the opposite side. Each segment cuts the angle perfectly in half and the 3 segments intersect at a single point.

Incenter theorem

The angle bisectors of a triangle intersect at a point that is equidistant from the sides of the triangle.

Three diagrams shown: left diagram titled inscrbed circle in an equilateral triangle, middle diagram titled inscribed circle in an isosceles triangle, and right triangle titled inscribed circle in a triangle. Speak to your teacher for more details.

Examples

Example 4

Find x. Justify your answer.

Worked Solution

Create a strategy

The vertical line segment is an angle bisector by the converse of angle bisector theorem which states that if a point is in the interior of an angle and is equidistant from the sides of the angle, then it lies on the angle bisector. Use the theorem to determine x.

Apply the idea

Since the vertical line segment is an angle bisector, x=23.

Example 5

P is the incenter of the triangle.

Triangle A B C with incenter P. Angle B has a measure of 76 degrees and angle C has a measure of 62 degrees.

Determine m \angle BAP.

Worked Solution

Create a strategy

Since P is the incenter of the triangle, it is the point of concurrency of the angle bisectors. So we know that \overline{AP} bisects \angle BAC, and therefore m\angle BAP = \dfrac{1}{2} m\angle BAC.

We can use the measures of the two known angles to find the measure of \angle BAC.

Apply the idea

\displaystyle m\angle BAC + m\angle ABC + m\angle ACB	\displaystyle =	\displaystyle 180\degree	Triangle interior angle sum
\displaystyle m\angle BAC + 76\degree + 62\degree	\displaystyle =	\displaystyle 180\degree	Substitute m\angle{ABC}=76\degree and m\angle{ACB}=62\degree
\displaystyle m\angle BAC +138\degree	\displaystyle =	\displaystyle 180\degree	Combine like terms
\displaystyle m\angle BAC	\displaystyle =	\displaystyle 42\degree	Subtract 138\degree from both sides

Then we have

\displaystyle m\angle BAP	\displaystyle =	\displaystyle \frac{1}{2} m\angle BAC	\overline{AP} bisects \angle BAC
\displaystyle m\angle BAP	\displaystyle =	\displaystyle \frac{1}{2} (42\degree)	Substitute m\angle{BAC}=42\degree
\displaystyle m\angle BAP	\displaystyle =	\displaystyle 21 \degree	Evaluate the multiplication

Determine m \angle BPC.

Worked Solution

Create a strategy

Again, since P is the incenter of the triangle, it is the point of concurrency of the angle bisectors. We can use this to find the measures of \angle PBC and \angle PCB. and then use those to determine the measure of \angle BPC.

Apply the idea

\displaystyle m\angle PBC	\displaystyle =	\displaystyle \frac{1}{2} m\angle ABC	Definition of angle bisector
	\displaystyle =	\displaystyle \frac{1}{2} \left(76 \degree\right)	Substitute m\angle{ABC}=76\degree
	\displaystyle =	\displaystyle 38 \degree	Evaluate the multiplication

and

\displaystyle m\angle PCB	\displaystyle =	\displaystyle \frac{1}{2} m\angle ACB	Definition of angle bisector
	\displaystyle =	\displaystyle \frac{1}{2} \left(62 \degree\right)	Substitute angle measure
	\displaystyle =	\displaystyle 31 \degree	Simplify expression

Then we have

\displaystyle m\angle BPC + m\angle PBC + m\angle PCB	\displaystyle =	\displaystyle 180\degree	Triangle interior angle sum
\displaystyle m\angle BPC + 38\degree + 31\degree	\displaystyle =	\displaystyle 180\degree	Substitute m\angle{BPC}=38\degree and m\angle{PCB}=31\degree
\displaystyle m\angle BPC + 69\degree	\displaystyle =	\displaystyle 180\degree	Combine like terms
\displaystyle m\angle BPC	\displaystyle =	\displaystyle 111\degree	Subtract 69\degree from both sides

Example 6

Draw a triangle and construct the inscribed circle.

Worked Solution

Create a strategy

The point of concurrency of the angle bisectors of a triangle is called the incenter because it is the center of the inscribed circle of the triangle.

To find the incenter, we need to draw at least two of the three angle bisectors. Two is sufficient as the three lines are concurrent, but three would be more precise if constructing by hand.

Once we have the incenter, we need to find the radius of the circle which will be the shortest (perpendicular) distance to the triangle, so we can draw a perpendicular line from the incenter to one of the sides and use that as the compass width for the inscribed circle.

Apply the idea

Draw a triangle and construct one of the angle bisectors.
Construct at least one more angle bisector.
Find the point of intersection of the two angle bisectors, this is the incenter. Then construct a line that is perpendicular to one of the sides of the triangle that goes through the incenter.
Set the compass width to the distance between the incenter and where the perpendicular line crossed the triangle side. Draw a circle with center at the incenter.
Erase all construction marks.

Example 7

A landscaper used a coordinate plane to plan the layout of a garden. The plans include the three paths shown below, that pass through the points A \left(30,0\right), B \left(30,45\right) and C \left(5,26.25\right). The landscaper wants to place a fountain at an equal distance from each path.

-5

x \left(\text{ft}\right)

-5

y \left(\text{ft}\right)

How could they find the coordinates of the fountain?

Worked Solution

Create a strategy

We need to identify whether perpendicular bisectors or angle bisectors will help us solve the problem. The goal in this problem is to find a location equal distance from each path.

Apply the idea

The landscaper can find the location of the fountain by considering the triangle formed by the paths and finding the intersection of the angle bisectors of the vertices of that triangle.

What is the location of the fountain?

Worked Solution

Create a strategy

Copy the diagram and use constructions to find the location of the fountain. We only need two angle bisectors to find the incenter of the paths.

Apply the idea

-5

x \left(\text{ft}\right)

-5

y \left(\text{ft}\right)

The fountain is located at X\left(20,25\right) on the coordinate plane.

Idea summary

The incenter is the point of concurrency of the angle bisectors of a triangle. It is called the incenter because it is the center of the inscribed circle of the triangle.

The incenter will be equidistant from each side of a triangle.

To construct the inscribed circle of a triangle, we can two construct angle bisectors on two angles and a perpendicular bisector. Then, we can draw a circle using the incenter and the point where the perpendicular line crosses the triangle's side as the radius.

Outcomes

G.CO.C.10

Prove theorems about triangles. Theorems include: measures of interior angles of a triangle sum to 180°; base angles of isosceles triangles are congruent; the segment joining midpoints of two sides of a triangle is parallel to the third side and half the length; the medians of a triangle meet at a point.

G.C.A.3

Construct the inscribed and circumscribed circles of a triangle, and prove properties of angles for a quadrilateral inscribed in a circle.

7.03 Angle and perpendicular bisectors

Introduction

Ideas

Perpendicular bisectors

Exploration

Examples

Example 1

Example 2

Example 3

Angle bisectors

Exploration

Examples

Example 4

Example 5

Example 6

Example 7

Outcomes

G.CO.C.10

G.C.A.3

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