Prove: Each median of an equilateral triangle divides the triangle into two congruent triangles.
Review Answers
Medians all have same length; distances from vertices to centroid – all are same; they are two-thirds the lengths of the medians.
Two of the medians have same length; distances from vertices to centroid are same for these two; all are two-thirds the lengths of the medians.
Medians all have different lengths; distances from vertices to centroid; all are different; they are two-thirds the lengths of the medians.
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False. The statement is true in the case of isosceles (vertex angle) and all angles in an equilateral triangle.
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The triangle is equilateral.
The difference in the areas of the inner and outer triangles is equal to the area of the original triangle.
The centroid will be located at . The midpoint of the vertical side of the triangle is located at . Note that is located units from point and that centroid will be one-third of the distance from point to .
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.
Using slopes and properties of straight lines, the point can be determined to have coordinates .
Use the distance formula to show that point is two-thirds of the way from each of the other two vertices to the midpoint of the opposite side.
Construct a median in an equilateral triangle. The triangles can be shown to be congruent by the SSS postulate.
Altitudes in Triangles
Learning Objectives
Construct the altitude of a triangle.
Apply the Concurrency of Altitudes Theorem to identify the point of concurrency of the altitudes of the triangle (the orthocenter).
Use the Concurrency of Altitudes Theorem to solve problems involving the orthocenter of triangles.
Introduction
In this lesson we will conclude our discussions about special line segments associated with triangles by examining altitudes of triangles. We will learn how to find the location of a point within the triangle that involves the altitudes.
Definition of Altitude of a Triangle
An altitude of a triangle is the line segment from a vertex perpendicular to the opposite side. Here is an example that shows the altitude from vertex in an acute triangle.
We need to be careful with altitudes because they do not always lie inside the triangle. For example, if the triangle is obtuse, then we can easily see how an altitude would lie outside of the triangle. Suppose that we wished to construct the altitude from vertex in the following obtuse triangle:
In order to do this, we must extend side as follows:
Will the remaining altitudes for (those from vertices and ) lie inside or outside of the triangle?
Answer: The altitude from vertex will lie inside the triangle; the altitude from vertex will lie outside the triangle.
As was true with perpendicular bisectors (which intersect at the circumcenter), and angle bisectors (which intersect at the incenter), and medians (which intersect at the centroid), we can state a theorem about the altitudes of a triangle.
Concurrency of Triangle Altitudes Theorem: The altitudes of a triangle will intersect in a point. We call this point the orthocenter of the triangle.
Rather than prove the theorem, we will demonstrate it for the three types of triangles (acute, obtuse, and right) and then illustrate some applications of the theorem.
Acute Triangles
The orthocenter lies within the triangle.
Obtuse Triangles
The orthocenter lies outside of the triangle.
Right Triangles
The legs of the triangle are altitudes. The orthocenter lies at the vertex of the right angle of the triangle.
Even with these three cases, we may still encounter special triangles that exhibit interesting properties.
Example 1
Use a piece of Patty Paper ( tracing paper), or any square piece of paper to explore orthocenters of isosceles . Note: Patty Paper may be purchased in bulk from many Internet sites.
Determine any relationships between the location of the orthocenter and the location of the incenter, circumcenter, and centroid.
First let’s recall that you can construct an isosceles triangle with Patty Paper as follows:
1. Draw line segment .
2. Fold point onto point to find the fold line.
3. Locate point anywhere on the fold line and connect point to points and . (Hint: Locate point as far away from and as possible so that you end up with a good-sized triangle.). Trace three copies of onto Patty Paper (so that you end up with four sheets of paper, each showing ).
4. For one of the sheets, fold the paper to locate the median, angle bisector, and perpendicular bisector relative to the vertex angle at point . What do you observe? (Answer: They are the same line segment.) Fold to find another bisector and locate the intersection of the two lines, the incenter.
5. For the second sheet, locate the circumcenter of .
6. For the third sheet, locate the centroid of .
7. For the third sheet, locate the orthocenter of .
8. Trace the location of the circumcenter, centroid, and orthocenter onto the original triangle. What do you observe about the four points? (Answer: The incenter, orthocenter, circumcenter, and centroids are collinear and lie on the median from the vertex angle.)
Do you think that the four points will be collinear for all other kinds of triangles? The answer is pretty interesting! In our homework we will construct the four points for a more general case.
Lesson Summary
In this lesson we:
Defined the orthocenter of a triangle.
Stated the Concurrency of Altitudes Theorem.
Solved problems using the Concurrency of Altitudes Theorem.
Examined the special case of an isosceles triangle and determined relationships about among the incenter, circumcenter, centroid, and orthocenter.
Points to Consider
Remember that the altitude of a triangle is also its height and can be used to find the area of the triangle. The altitude is the shortest distance from a vertex to the opposite side.
Review Questions
In our lesson we looked at the special case of an isosceles triangle and determined relationships about among the incenter, circumcenter, centroid, and orthocenter. Explore the case of an equilateral triangle and see which (if any) relationships hold.
Perform the same exploration for an acute triangle. What can you conclude?
Perform the same exploration for an obtuse triangle. What can you conclude?
Perform the same exploration for a right triangle. What can you conclude?
What can you conclude about the four points for the general case of ?
In 3 you found that three of the four points were collinear. The segment joining these three points define the Euler segment. Replicate the exploration of the general triangle case and measure the lengths of the Euler segment and the sub-segments. Drag your drawing so that you can investigate potential relationships for several different triangles. What can you conclude about the lengths?
(Found in Exploring Geometry, 1999, Key Curriculum Press) Construct a triangle and find the Euler segment. Construct a circle centered at the midpoint of the Euler segment and passing through the midpoint of one of the sides of the triangle. This circle is called the nine-point circle. The midpoint it passes through is one of the nine points. What are the other eight?
Consider with, , altitudes of the triangles as indicated.
Prove: .
Consider isosceles triangle with , and . Prove: .
Review Answers
All four points are the same.
The four points all lie inside the triangle.
The four poin
ts all lie outside the triangle.
The orthocenter lies on the vertex of the right angle and the circumcenter lies on the midpoint of the hypotenuse.
The orthocenter, the circumcenter, and the centroid are always collinear.
The circumcenter and the orthocenter are the endpoints of the Euler segment.
The distance from the orthocenter to the centroid is twice the distance from the centroid to the circumcenter.
Three of the points are the midpoints of the triangle’s sides. Three other points are the points where the altitudes intersect the opposite sides of the triangle. The last three points are the midpoints of the segments connecting the orthocenter with each vertex.
The congruence can be proven by showing the congruence of triangles and . This can be done by applying postulate AAS to the two triangles.
The proof can be completed by using the AAS postulate to show that triangles and are congruent. The conclusion follows from CPCTC.
Inequalities in Triangles
Learning Objectives
Determine relationships among the angles and sides of a triangle.
Apply the Triangle Inequality Theorem to solve problems.
Introduction
In this lesson we will examine the various relationships among the measure of the angles and the lengths of the sides of triangles. We will do so by stating and proving a few key theorems that will enable us to determine the types of relationships that hold true.
Look at the following two triangles
We see that the first triangle is isosceles while in the the second triangle, is longer than . How are the measures of the angles at and related to the lengths of and It appears (and, in fact it is the case) that the measure of the angle at vertex is larger than .
In this section we will formally prove theorems that reveal when such relationships hold. We will start with the following theorem.
Relationship Between the Sides and Angles of a Triangle
Theorem: If two sides of a triangle are of unequal length, then the angles opposite these sides are also unequal. The larger side will have a larger angle opposite it.
Proof. Consider with . We must show that .
1. By the Ruler Postulate, there is a point on such that . Construct and label angles , and as follows.
2. Since is isosceles, we have .
3. By angle addition, we have .
4. So , and by substitution .
5. Note that is exterior to , so .
6. Hence, and , we have .
We can also prove a similar theorem about angles.
Larger angle has longer opposite side: If one angle of a triangle has greater measure than a second angle, then the side opposite the first angle is longer than the side opposite the second angle.
Proof. In order to prove the theorem, we will use a method that relies on indirect reasoning, a method that we will explore further. The method relies on starting with the assumption that the conclusion of the theorem is wrong, and then reaching a conclusion that logically contradicts the given statements.
1. Consider with . We must show that .
2. Assume temporarily that is not greater than . Then either or .
3. If , then the angles at vertices and are congruent. This is a contradiction of our given statements.
4. If , then by the fact that the longer side is opposite the largest angle (the theorem we just proved). But this too contradicts our given statements. Hence, we must have .
With these theorems we can now prove an interesting corollary.
Corollary The perpendicular segment from a point to a line is the shortest segment from the point to the line.
Proof. The proof is routine now that we have proved the major results.
Consider point and line and the perpendicular line segment from to as follows.
We can draw the segment from to any point on line and get the case of a right triangle as follows:
Since the triangle is a right triangle, the side opposite the right angle (the hypotenuse) will always have length greater than the length of the perpendicular segment from to , which is opposite an angle of .
Now we are ready to prove one of the most useful facts in geometry, the triangle inequality theorem.
Triangle Inequality Theorem: The sum of the lengths of any two sides of a triangle is greater than the length of the third side.
Proof. Consider . We must show the following:
Suppose that is the longest side. Then statements 2 and 3 above are true.
In order to prove 1. , construct the perpendicular from point to on the opposite side as follows:
Now we have two right triangles and can draw the following conclusions:
Since the perpendicular segment is the shortest path from a point to a line (or segment), we have is the shortest segment from to . Also, is the shortest segment from to . Therefore and and by addition we have
So, .
Example 1
Can you have a triangle with sides having lengths ?
Without a drawing we can still answer this question—it is an impossible situation, we cannot have such a triangle. By the Triangle Inequality Theorem, we must have that the sum of lengths of any two sides of the triangle must be greater than the length of the third side. In this case, we note that .
Example 2
Find the angle of smallest measure in the following triangle.
has the smallest measure. Since the triangle is a right triangle, we can find using the Pythagorean Theorem (which we will prove later).
By the fact that the longest side is opposite the largest angle in a triangle, we can conclude that .
Lesson Summary
In this lesson we:
Stated and proved theorems that helped us determine relationships among the angles and sides of a triangle.
Introduced the method of indirect proof.
Applied the Triangle Inequality Theorem to solve problems.
Points to Consider
Knowing these theorems and the relationships between the angles and sides of triangles will be applied when we use trigonometry. Since the size of the angle affects the length of the opposite side, we can show that there are specific angles associated with certain relationships (ratios) between the sides in a right triangle, and vice versa.
Review Questions
Name the largest and smallest angles in the following triangles:
Name the longest side and the shortest side of the triangles.
Is it possible to have triangles with the following lengths? Give a reason for your answer.
Two sides of a triangle have lengths and . What can you conclude about the length of the third side?
The base of an isosceles triangle has length . What can you say about the length of each leg?
In exercises 6 and 7, find the numbered angle that has the largest measure of the triangle.
In exercises 8-9, find the longest segment in the diagram.
Given:
Prove:
Given:
Prove:
Review Answers
is largest and is smallest.
is largest and is smallest.
is longest and is the shortest.
is longest and is the shortest.
No, .
Yes
No, .
Yes
The third side must have length such that .
The legs each must have length greater than .
Since the angle opposite each of the two segments that comprise is greater than the angle opposite the corresponding segments of .
Proof
Inequalities in Two Triangles
Learning Objectives
Determine relationships among the angles and sides of two triangles.
Apply the SAS and SSS Triangle Inequality Theorems to solve problems.
Introduction
In our last lesson we examined the various relationships among the measure of angles and the lengths of the sides of triangles and proved
the Triangle Inequality Theorem that states that the sum of the lengths of two sides of a triangle is greater than the third side. In this lesson we will look at relationships in two triangles.
SAS Inequality Theorem
Let’s begin our discussion by looking at the following congruent triangles.
If we think of the sides of the triangle as matchsticks that are “hinged” at and respectively then we can increase the measure of the angles by opening up the sticks. If we open them so that , then we see that . Conversely, if you open them so that , then we see that . We can prove theorems that involve these relationships.
SAS Inequality Theorem (The Hinge Theorem): If two sides of a triangle are congruent to two sides of another triangle, but the included angle of the first triangle has greater measure than the included angle of the second triangle, then the third side of the first triangle is longer than the third side of the second triangle.
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