Construct the perpendicular bisector of . Note that the slope of is .
Point will be located on the perpendicular bisector. The perpendicular bisector will have slope .
The perpendicular bisector will pass through the midpoint and have slope Its equation is
So, the distance from to is equal to , which is .
Consider the distance from point to . Solve the following distance equation to find the coordinate of point .
Since lies on the line , use the value of found from the equation to find the coordinate.
Show that use CPCTC and the definitions of bisector and properties about congruent adjacent angles forming a straight angle.
Angle Bisectors in Triangles
Learning Objectives
Construct the bisector of an angle.
Apply the Angle Bisector Theorem to identify the point of concurrency of the perpendicular bisectors of the sides (the incenter).
Use the Angle Bisector Theorem to solve problems involving the incenter of triangles.
Introduction
In our last lesson we examined perpendicular bisectors of the sides of triangles. We found that we were able to use perpendicular bisectors to circumscribe triangles. In this lesson we will learn how to inscribe circles in triangles. In order to do this, we need to consider the angle bisectors of the triangle. The bisector of an angle is the ray that divides the angle into two congruent angles.
Here is an example of an angle bisector in an equilateral triangle.
Angle Bisector Theorem and its Converse
We can prove the following pair of theorems about angle bisectors.
Angle Bisector Theorem: If a point is on the bisector of an angle, then the point is equidistant from the sides of the angle.
Before we proceed with the proof, let’s recall the definition of the distance from a point to a line. The distance from a point to a line is the length of the line segment that passes through the point and is perpendicular to the original line.
Proof. Consider with angle bisector , and segments and , perpendicular to each side through point as follows:
We will show that .
Since is the bisector of , then by the definition of angle bisector. In addition, since and are perpendicular to the sides of , then and are right angles and thus congruent. Finally, by the reflexive property.
By the AAS postulate, we have .
So by CPCTC (corresponding parts of congruent triangles are congruent).
Therefore is equidistant from each side of the angle. And since represents any point on the angle bisector, we can say that every point on the angle bisector is equidistant from the sides of the angle.
We can also prove the converse of this theorem.
Converse of the Angle Bisector Theorem: If a point is in the interior of an angle and equidistant from the sides, then it lies on the bisector of the angle.
Proof. Consider with points and and segment such that as follows:
As the distance to each side is given by the lengths of and respectively, we have that and are perpendicular to sides and respectively.
Note that is the hypotenuse of right triangles and Hence, since , and , and and are right angles, then the triangles are congruent by Theorem 4-6.
by CPCTC.
Hence, point lies on the angle bisector of .
Notice that we just proved the Angle Bisector Theorem (If a point is on the angle bisector then it is equidistant from the sides of the angle) and we also proved the converse of the Angle Bisector theorem (If a point is equidistant from the sides of an angle then it is on the angle bisector of the triangle). When we have proven both a theorem and its converse we say that we have proven a biconditional statement. We can put the two conditional statements together using if and only if: "A point is on the angle bisector of an angle if and only if it is equidistant from the sides of the triangle."
Angle Bisectors in a Triangle
We will now use these theorems to prove an interesting result about the angle bisectors of a triangle.
Concurrency of Angle Bisectors Theorem: The angle bisectors of a triangle intersect in a point that is equidistant from the three sides of the triangle.
Proof. We will use the previous two theorems to establish the proof.
1. Consider .
2. We can construct the angle bisectors of and intersecting at point as follows.
3. We will show that point is equidistant from sides , , and and that is on the bisector of .
4. Construct perpendicular line segments from point to sides , and as follows:
5. Since is on the bisectors of and , then by Theorem 5-5, . Therefore, is equidistant from sides , and .
6. Since is equidistant from and , Theorem 5-6 applies and we must have that is on the angle bisector of .
The point has a special property. Since it is equidistant from each side of the triangle, we can see that is the center of a circle that lies within the triangle. We say that the circle is inscribed within the triangle and the point is called the incenter of the triangle. This is illustrated in the following figure.
Example 1
Inscribe the following triangle using a compass and a straightedge.
1. Draw triangle with your straightedge.
2. Use your compass to construct the angle bisectors and find the point of concurrency .
3. Use your compass to construct the circle that inscribes .
Example 2
Inscribe a circle within the following triangle using The Geometer’s Sketchpad.
We can use the commands of GSP to construct the incenter and corresponding circle as follows:
1. Open a new sketch and construct triangle using the Segment Tool.
2. You can construct the angle bisectors of the angles by first designating the angle by selecting the appropriate vertices (e.g., to select the angle at vertex , select points , and in order) and then choosing Construct Angle Bisector from the Construct menu. After bisecting two angles, construct the point of intersection by selecting each angle bisector and choosing Intersection from the Construct menu. (Recall from our proof of the concurrency of angle bisectors theorem that we only need to bisect two of the angles to find the incenter.)
3. You are now ready to construct the circle. Recall that the radius of the circle must be the distance from to each side – our figure above does not include that segment. However, we do not need to construct the perpendicular line segments as we did to prove Theorem 5-7. Sketchpad will measure the distance for us.
4. To measure the distance from to each side, select point and one side of the triangle. Choose Distance from the Measure menu. This will give you the radius of your circle.
5. We are now ready to construct the circle. Select point and the distance from to the side of the triangle. Select “Construct circle by center + radius” from the Construct menu. This will give the inscribed circle within the triangle.
Lesson Summary
In this lesson we:
Defined the angle bisector of an angle.
Stated and proved the Angle Bisector Theorem.
Solved problems using the Angle Bisector Theorem.
Constructed angle bisectors and the inscribed circle with compass and straightedge, and with Geometer’s Sketchpad.
Points to Consider
How are circles related to triangles, and how are triangles related to circles? If we draw a circle first, what are the possibilities for the triangles we can circumscribe? In later chapters we will more carefully define and work with the properties of circles.
Review Questions
Construct the incenter of and the inscribed circle for each of the following triangles using a straightedge, compass, and Geometer's Sketchpad.
In the last lesson we found that we could circumscribe some kinds of quadrilaterals as long as opposite angles were supplementary. Use Geometer's Sketchpad to explore the following quadrilaterals and see if you can inscribe them by the angle bisector method. a square
a rectangle
a parallelogram
a rhombus
From your work in a-d, what condition must hold in order to circumscribe a quadrilateral?
Consider equilateral triangle . Construct the angle bisectors of the triangle and the incenter . Connect the incenter to each vertex so that the line segment intersects the side opposite the angle as follows.
As with circumcenters, we get six congruent triangles. Now connect the points that intersect the sides. What kind of figure do you get?
True or false: An incenter can also be a circumcenter. Illustrate your reasoning with a drawing.
Consider the situation described in exercise 4 for the case of an isosceles triangle. What can you conclude about the six triangles that are formed?
Consider line segment with coordinates , . Suppose that we wish to find points and so that the resulting quadrilateral can be either circumscribed or inscribed. What are some possibilities for locating points and
Using a piece of tracing or Patty Paper, construct an equilateral triangle. Bisect one angle by folding one side onto another. Unfold the paper. What can you conclude about the fold line?
Repeat exercise 8 with an isosceles triangle. What can you conclude about repeating the folds?
What are some other kinds of polygons where you could use Patty Paper to bisect an angle into congruent figures?
Given: is the perpendicular bisector of .
is the perpendicular bisector of .
Prove:
.
Given: bisects .
bisects .
Prove:
bisects .
Review Answers
Yes
No: Bisectors are not concurrent at a point.
No: Bisectors are not concurrent at a point.
Yes
Angle bisectors must be concurrent.
Equilateral triangle
The statement is true in the case of an equilateral triangle. In addition, for squares the statement is also true.
We do not get six congruent triangles as before. But we get four congruent triangles and a separate pair of congruent triangles. In addition, if we connect the points where the bisectors intersect the sides, we get an isosceles triangle.
From our previous exercises we saw that we could inscribe and circumscribe some but not all types of quadrilaterals. Drawing from those exercises, we see that we could circumscribe and inscribe a square. So, locating the points at and is one such possibility. Similarly, we could locate the points at and and get a kite that can be inscribed but not circumscribed.
The fold line divides the triangle into two congruent triangles and thus is a line of symmetry for the triangle. Note that the same property will hold by folding at each of the remaining angles.
There is only one fold line that divides the triangle into congruent triangles, the line that folds the angle formed by the congruent sides.
Any regular polygon will have this property. For example, a regular pentagon:
Medians in Triangles
Learning Objectives
Construct the medians of a triangle.
Apply the Concurrency of Medians Theorem to identify the point of concurrency of the medians of the triangle (the centroid).
Use the Concurrency of Medians Theorem to solve problems involving the centroid of triangles.
Introduction
In our two last lessons we learned to circumscribe circles about triangles by finding the perpendicular bisectors of the sides and to inscribe circles within triangles by finding the triangle’s angle bisectors. In this lesson we will learn how to find the location of a point within the triangle that involves the medians.
Definition of Median of a Triangle
A median of a triangle is the line segment that joins a vertex to the midpoint of the opposite side.
Here is an example that shows the medians in an obtuse triangle.
That the three medians appear to intersect in a point is no coincidence. As was true with perpendicular bisectors of the triangle sides and with angle bisectors, the three medians will be concurrent (intersect in a point). We call this point the centroid of the triangle. We can prove the following theorem about centroids.
The Centroid of a Triangle
Concurrency of Medians Theorem: The medians of a triangle will intersect in a point that is two-thirds of the distance from the vertices to the midpoint of the opposite side.
Consider with midpoints of the sides located at , , and and the point of concurrency of the medians at the centroid, . The theorem states that , and .
The theorem can be proved using a coordinate system and the midpoint and distance formulas for line segments. We will leave the proof to you (Homework Exercise #10), but will provide an outline and helpful hints for developing the proof.
Example 1.
Use The Concurrency of Medians Theorem to find the lengths of the indicated segments in the following triangle that has medians , and as indicated.
1. If , then ____ and ____.
2. If . then ____ and _____.
We will start by finding .
Now for ,
Napoleon's Theorem
In the remainder of the lesson we will provide an interesting application of a theorem attributed to Napoleon Bonaparte, Emperor of France from 1804 to 1821, which makes use of equilateral triangles and centroids. We will explore Napoleon’s theorem using The Geometer’s Sketchpad.
But first we need to review how to construct an equilateral triangle using circles. Consider and circles having equal radius and centered at and as follows:
Once you have hidden the circles, you will have an equilateral triangle. You can use the construction any time you need to construct an equilateral triangle by selecting the finished triangle and then making a Tool using the tool menu.
Preliminary construction for Napoleon’s Theorem: Construct any triangle . Construct an equilateral triangle on each side.
Find the centroid of each equilateral triangle and connect the centroids to get the Napoleon outer triangle.
Measure the sides of the new triangle using Sketchpad. What can you conclude about the Napoleon outer triangle? (Answer: The triangle is equilateral.)
This result is all the more remarkable since it applies to any triangle . You can verify this fact in GSP by "dragging" a vertex of the original triangle to form other triangles. The outer triangle will remain equilateral. Homework problem 9 will allow you to further explore this theorem.
Example 2
Try this:
Draw a triangle on a sheet of card stock paper (or thin cardboard) and locate the centroid.
Carefully cut out the triangle.
Hold your pencil point up and place the triangle on it so that the centroid rests on the pencil.
What do you notice?
The triangle balances on the pencil. Why does the triangle balance?
Lesson Summary
In this lesson we:
Defined the centroid of a triangle.
Stated and proved the Concurrency of Medians Theorem.
Solved problems using the Concurrency of Medians Theorem.
Demonstrated Napoleon’s Theorem.
Points to Consider
So far we have been looking at relationships within triangles. In later chapters we will review the area of a triangle. When we draw the medians of the triangle, six smaller triangles are created. Think about the area of these triangles, and how that might relate to example 1 above.
Review Questions
Find the centroid of for each of the following triangles using Geometer's Sketchpad. For each triangle, measure the lengths of the medians and the distances from the centroid to each of the vertices. What can you conclude for each of the triangles? an equilateral triangle
an isosceles triangle
A scalene triangle
has points as midpoints of sides and the centroid located at point as follows.
Find the following lengths if ____
, ___.
True or false: A median cannot be an angle bisector. Illustrate your reasoning with a drawing.
Find the coordinates of the centroid of with vertices , and .
Find the coordinates of the centroid of with vertices and . Also, find .
Use the example sketch of Napoleon’s Theorem to form the following triangle: Reflect each of the centroids in the line that is the closest side of the original triangle.
Connect the points to form a new triangle that is called the inner Napoleon triangle.
What can you conclude about the inner Napoleon triangle?
You have been asked to design a triangular metal logo for a club at school. Using the following rectangular coordinates, determine the logo’s centroid.
Prove Theorem 5-8. An outline of the proof together with some helpful hints is provided here. Proof. Consider with as follows:
Hints: Note that the midpoint of side is located at the origin. Construct the median from vertex to the origin, and call it . The point of concurrency of the three medians will be located on at point that is two-thirds of the way from to the origin.
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