Use the following diagram of for exercise 1-4. Given the coordinates and find the new coordinates of after each transformation.
Slide down three units.
Slide up and to the right .
Reflect across the axis.
Rotate clockwise about the origin. Draw a sketch to help visualize what this looks like.
Use the following diagram that shows a transformation of to for exercises 5-7:
What kind of transformation was used to go from to ?
Use the distance formula to show that
Is the transformation congruence preserving? Justify your answer.
Use the following diagram for exercises 8-9.
What kind of transformation is shown above?
Is this transformation congruence preserving? Justify your answer.
Can a rotation be described in terms of reflections? Justify your answer.
Review Answers
Reflection about the axis
.
This is a dilation.
No, we can see that each side of the larger triangle is twice as long as the corresponding side in the original, so this is not length preserving.
Yes, a rotation about the origin is the same as two reflections done consecutively, one across the axis and then one across the axis. For a ; rotation, the rule for transforming coordinates is . Now, suppose point is reflected twice, first across the axis, and then across the axis. After the first transformation, the coordinates are Then after reflection on the axis, we get , which is the same coordinates that result from a rotation.
Chapter 5: Relationships Within Triangles
Midsegments of a triangle
Learning Objectives
Identify the midsegment of a triangle.
Apply the Midsegment Theorem to solve problems involving side lengths and midsegments of triangles.
Use the Midsegment Theorem to solve problems involving variable side lengths and midsegments of triangles.
Introduction
In previous lessons, we used the parallel postulate to learn new theorems that enabled us to solve a variety of problems about parallel lines:
Parallel Postulate: Given: line and a point not on . There is exactly one line through that is parallel to .
In this lesson we extend these results to learn about special line segments within triangles. For example, the following triangle contains such a configuration:
Triangle is cut by where and are midpoints of sides and respectively. is called a midsegment of . Note that has other midsegments in addition to . Can you see where they are in the figure above?
If we construct the midpoint of side at point and construct and respectively, we have the following figure and see that segments and are midsegments of .
In this lesson we will investigate properties of these segments and solve a variety of problems.
Properties of midsegments within triangles
We start with a theorem that we will use to solve problems that involve midsegments of triangles.
Midsegment Theorem: The segment that joins the midpoints of a pair of sides of a triangle is:
parallel to the third side.
half as long as the third side.
Proof of 1. We need to show that a midsegment is parallel to the third side. We will do this using the Parallel Postulate.
Consider the following triangle . Construct the midpoint of side .
By the Parallel Postulate, there is exactly one line though that is parallel to side . Let’s say that it intersects side at point . We will show that must be the midpoint of and then we can conclude that is a midsegment of the triangle and is parallel to .
We must show that the line through and parallel to side will intersect side at its midpoint. If a parallel line cuts off congruent segments on one transversal, then it cuts off congruent segments on every transversal. This ensures that point is the midpoint of side .
Since , we have . Hence, by the definition of midpoint, point is the midpoint of side . is a midsegment of the triangle and is also parallel to .
Proof of 2. We must show that .
In , construct the midpoint of side at point and midsegments and as follows:
First note that by part one of the theorem. Since and , then and since alternate interior angles are congruent. In addition, .
Hence, by The ASA Congruence Postulate. since corresponding parts of congruent triangles are congruent. Since is the midpoint of , we have and by segment addition and substitution.
So, and .
Example 1
Use the Midsegment Theorem to solve for the lengths of the midsegments given in the following figure.
, and are midpoints of the sides of the triangle with lengths as indicated. Use the Midsegment Theorem to find
B. The perimeter of the triangle .
A. .
B. By the Midsegment Theorem, implies that ; similarly, , and . Hence, the perimeter is
A. Since is a midpoint, we have and . By the theorem, we must have .
We can also examine triangles where one or more of the sides are unknown.
Example 2
Use the Midsegment Theorem to find the value of in the following triangle having lengths as indicated and midsegment .
By the Midsegment Theorem we have . Solving for , we have .
Lesson Summary
In this lesson we:
Introduced the definition of the midsegment of a triangle and examined examples.
Stated and proved the Midsegment Theorem.
Solved problems using the Midsegment Theorem.
Review Questions
are midpoints of sides of triangles and .
Complete the following:
If , then ___ and ___.
If , then ____.
If and , then ___and ___.
If and , then _____.
Consider triangle with vertices and midpoint on .
Find the coordinates of point .
Use the Midsegment Theorem to find the coordinates of the point on side that makes the midsegment.
For problem 5, describe another way to find the coordinates of point that does not use the Midsegment Theorem.
In problems 7-8, the segments join the midpoints of two sides of the triangle. Find the values of and for each problem.
In triangle , sides , and have lengths and respectively. Triangle is formed by joining the midpoints of . Find the perimeter of .
For the original triangle of 9, find its perimeter and compare to the perimeter of .
Can you state a relationship between a triangle’s perimeter and the perimeter of the triangle formed by connecting its midsegments?
Given: is the midpoint of . Prove: .
Given: is the midpoint of , , and .
Can you conclude that If true, prove the assertion. If false, provide a counterexample.
Review Answers
and
Find midpoint and then the slope of . Find the line through parallel to (line ). Find the equation of the line that includes (line ). Find the intersection of lines and .
,
,
The perimeter of is The perimeter of is
The perimeter of the midsegment triangle will always be half the perimeter of the original triangle.
Use the givens and Theorem 5-1 to show that point is the midpoint of .
The assertion is true. Using Theorem 5-1, it can be shown that the triangles are congruent by SSS postulate.
Perpendicular Bisectors in Triangles
Learning Objectives
Construct the perpendicular bisector of a line segment.
Apply the Perpendicular Bisector Theorem to identify the point of concurrency of the perpendicular bisectors of the sides (the circumcenter).
Use the Perpendicular Bisector Theorem to solve problems involving the circumcenter of triangles.
Introduction
In our last lesson we examined midsegments of triangles. In this lesson we will examine another construction that can o
ccur within triangles, called perpendicular bisectors.
The perpendicular bisector of a line segment is the line that:
divides the line segment into two congruent sub-segments.
intersects the line segment at a right angle.
Here is an example of a perpendicular bisector to line segment .
Perpendicular Bisector Theorem and its Converse
We can prove the following pair of theorems about perpendicular bisectors.
Perpendicular Bisector Theorem: If a point is on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.
Proof. Consider with perpendicular bisector with points and on line as follows:
We must show that .
Since is the perpendicular bisector of , it follows that and angles and are congruent and are right angles.
By the SAS postulate, we have .
So by CPCTC (corresponding parts of congruent triangles are congruent).
It turns out that we can also prove the converse of this theorem.
Converse of the Perpendicular Bisector Theorem: If a point is equidistant from the endpoints of a segment, then the point is on the perpendicular bisector of the segment.
Proof. Consider as follows with .
We will construct the midpoint of and show that is the perpendicular bisector to .
1. Construct the midpoint of at point Construct .
2. Consider and . These are congruent triangles by postulate SSS.
3. So by CPCTC, we have .
4. Since and form a straight angle and are also congruent, then . Hence, is on the perpendicular bisector to .
Notice that we just proved the Perpendicular Bisector Theorem and we also proved the Converse of the Perpendicular Bisector Theorem. When you prove a theorem and its converse you have proven a biconditional statement. We can state the Perpendicular Bisector Theorem and its converse in one step: A point is on the perpendicular bisector of a segment if and only if that point is equidistant from the endpoints of the segment.
We will now use these theorems to prove an interesting result about the perpendicular bisectors of the sides of a triangle.
Concurrency of Perpendicular Bisectors: The perpendicular bisectors of the sides of a triangle intersect in a point that is equidistant from the vertices.
Proof. We will use the previous two theorems to establish the proof.
1. Consider
2. We can construct the perpendicular bisectors of sides and intersecting at point as follows.
3. We will show that point also lies on the perpendicular bisector of and thus is equidistant from the vertices , and .
4. Construct line segments and as follows.
5. Since is on the perpendicular bisector of , then is equidistant from and by the Perpendicular Bisector Theorem and . Similarly, is on the perpendicular bisector of , then is equidistant from and by the Perpendicular Bisector Theorem. Therefore, .
6. By the transitive law, we have . By the Converse of the Perpendicular Bisector Theorem, we must have that is on the perpendicular bisector of .
The point has a special property. Since it is equidistant from each vertex, we can see that is the center of a circle that circumscribes the triangle. We call the circumcenter of the triangle. This is illustrated in the following figure.
Example 1
Construct a circumscribed triangle using a compass and a straightedge.
1. Draw triangle with your straightedge.
2. Use your compass to construct the perpendicular bisectors of the sides and find the point of concurrency .
3. Use your compass to verify that .
4. Use your compass to construct the circle that circumscribes
Example 2
Construct a circumscribed triangle using The Geometer’s Sketchpad (GSP)
We can use the commands of GSP to construct the circumcenter and corresponding circle as follows.
1. Open a new sketch and construct triangle using the Segment Tool.
2. You can construct the perpendicular bisectors of the sides by going to the Construct menu and choosing the following options. Select each side and choose Construct Midpoints. Then for each side select the midpoint and the side (and nothing else), and then choose Construct Perpendicular Bisector.
3. Select two of the three bisectors and choose Construct Point of Intersection from the Construct menu. This will provide point , the circumcenter.
4. Construct the circle having center and passing through points and . Recall that there are two ways to construct the circle: . Using the draw tool on the left column, and . Using the Construct Menu. For this construction, you will want to use the Construct menu to ensure that the circle passes through the vertices.
As a further exploration, try the following with paper:
Cut out any triangle from a sheet of paper.
Fold the the triangle over one side so that the side is folded in half.
Repeat for the other two sides.
What do you notice?
Notice that the folds will cross at the circumcenter, unless the triangle is obtuse. In which case the fold lines will meet outside the triangle if they continued.
Lesson Summary
In this lesson we:
Defined the perpendicular bisector of a line segment.
Stated and proved the Perpendicular Bisector Theorem.
Solved problems using the Perpendicular Bisector Theorem.
Points to Consider
If we think about three non-collinear points in a plane, we can imagine a triangle that has each point as a vertex. Locating the circumcenter, we can draw a circle that all three vertices will be on. What does this tell us about any three non-collinear points in a plane?
There is a unique circle for any three non-collinear points in the same plane.
Finding a circle through any three points will also work in coordinate geometry. You can use the circumcenter to find the equation of a circle through any three points. In calculus this method is used (together with some tools that you have probably not learned yet) to precisely describe the curvature of any curve.
Review Questions
Construct the circumcenter of and the circumscribed circle for each of the following triangles using a straightedge, compass, and Geometer's Sketchpad.
Based on your constructions in 1-3, state a conjecture about the relationship between a triangle and the location of its circumcenter.
In this lesson we found that we could circumscribe a triangle by finding the point of concurrency of the perpendicular bisectors of each side. Use Geometer's Sketchpad to see if the method can be used to circumscribe each of the following figures: a square
a rectangle
a parallelogram
From your work in a-c, what condition must hold in order to circumscribe a quadrilateral?
Consider equilateral triangle . Construct the perpendicular bisectors of the sides of the triangle and the circumcenter . Connect the circumcenter to each vertex. Your original triangle is now divided into six triangles. What can you conclude about the six triangles?
Suppose three cities and are situated as follows.
The leaders of these cities wish to construct a new health center that is equidistant from each city. Is this a wise plan? Why or why not?
True or false: An isosceles triangle will always have its circumcenter located inside the triangle? Give reasons for your answer.
True or false: The perpendicular bisectors of an equilateral triangle intersect in the exact center of the triangle’s interior. Give reasons for your answer.
Consider line segment with coordinates . Suppose that we wish to find point so that is equilateral. How can you use perpendicular bisectors to find the location of point ?
Suppose that is a right triangle as indicated:
Construct the bisector of .
Prove: is the perpendicular bisector of .
Review Answers
If triangle is acute, then
the circumcenter lies inside of the triangle. If triangle is obtuse, then the circumcenter lies outside of the triangle. If triangle is a right triangle, then the circumcenter lies on the hypotenuse of the triangle.
Yes
Yes
No
Opposite angles must be supplementary.
The triangles are congruent to one another and each is a right triangle.
It is not a wise plan. Since and form an obtuse triangle, the location of the circumcenter would be outside the triangle. Hence, the health center would be located at the circumcenter, which would be a much greater distance from each city than the distance between the cities themselves.
False. It is possible to have isosceles triangles that are acute, obtuse, and right. Hence, there are isosceles triangles where the circumcenter could be located outside the triangle (in the case of an obtuse triangle) or on the boundary of the triangle (in the case of a right triangle).
True. See the solution to problem 10 to verify this fact.
CK-12 Geometry Page 18