Bisectors of Triangles
As you well know by now, being able to deduce key information from a limited set of facts is the basis of geometry. An important type of segment, ray, or line that can help us prove congruence is called an angle bisector. Understanding what angle bisectors are and how they affect triangle relationships is crucial as we continue our study of geometry. Let's investigate different types of bisectors and the theorems that accompany them.
A perpendicular bisector is a special kind of segment, ray, or line that
(1) intersects a given segment at a 90° angle, and
(2) passes through the given segment's midpoint.
Segment CD is the perpendicular bisector to segment AB.
We derive two important theorems from the characteristics of perpendicular bisectors. We can use these theorems in our two-column geometric proofs, or we can just use them to help us in geometric computations.
Perpendicular Bisector Theorem
If a point lies on the perpendicular bisector of a segment, then it is equidistant from the endpoints of the segment.
Converse also true: If a point is equidistant from the endpoints of a segment, then it lies on the perpendicular bisector of the segment.
These theorems essentially just show that there exist a locus of points (which form the perpendicular bisector) that are equidistant from the endpoints of a given segment which meet at the midpoint of the segment at a right angle. An illustration of this concept is shown below.
Points E, F, G, and H (along with an infinite amount of points) are equidistant from A and B. Together, they form the perpendicular bisector of segment AB.
The perpendicular bisectors of a triangle have a very special property. Let's investigate it right now.
The perpendicular bisectors of the sides of a triangle intersect at a point called the circumcenter of the triangle, which is equidistant from the vertices of the triangle.
Point G is the circumcenter of ?ABC.
Now, we will study a geometric concept that will help us prove congruence between two angles. Any segment, ray, or line that divides an angle into two congruent angles is called an angle bisector.
We will use the following angle bisector theorems to derive important information from relatively simple geometric figures.
Angle Bisector Theorem
If a point lies on the bisector of an angle, then it is equidistant from the sides of the angle.
Converse also true: If a point in the interior of an angle is equidistant from the sides of the angle, then it lies on the bisector of the angle.
The points along ray AD are equidistant from either side of the angle. Together, they form a line that is the angle bisector.
Similar to the perpendicular bisectors of a triangle, there is a common point at which the angle bisectors of a triangle meet. Let's look at the corresponding theorem.
The angle bisectors of a triangle intersect at a point called the incenter of the triangle, which is equidistant from the sides of the triangle.
Point G is the incenter of ?ABC.
While similar in many respects, it will be important to distinguish between perpendicular bisectors and angle bisectors. We use perpendicular bisectors to create a right angle at the midpoint of a segment. Any point on the perpendicular bisector is equidistant from the endpoints of the given segment. The point at which the perpendicular bisectors of a triangle meet, or the circumcenter, is equidistant from the vertices of the triangle.
On the other hand, angle bisectors simply split one angle into two congruent angles. Points on angle bisectors are equidistant from the sides of the given angle. We also note that the points at which angle bisectors meet, or the incenter of a triangle, is equidistant from the sides of the triangle.
Let's work on some exercises that will allow us to put what we've learned about perpendicular bisectors and angle bisectors to practice.
BC is the perpendicular bisector of AD. Find the value of x.
The most important fact to notice is that BC is the perpendicular bisector of AD because, although it is just one statement, we can derive much information about the figure from it. The fact that it is a perpendicular bisector implies that segment DB is equal to segment AB since it passes through the midpoint of segment AD. Therefore, we have
Subtracting 7x from both sides of the equation yields
So, we have x=6.
N is the circumcenter of ?RAK. Find the values of x and y.
We are aware that segments RM, XS, and YE are all perpendicular bisectors since they meet at N, the circumcenter of ?RAK. Thus, in order to solve for x, we can set segments RE and AE equal to each other since E is the midpoint of segment RA. We have
In order to solve for y, we have to use the information given by the Circumcenter Theorem. This theorem states that the circumcenter is equidistant from the vertices of the triangle. Thus, we know that RN=AN=KN. For this part of the problem, we only need to solve for y with AN=KN. We have
So, we have x=4 and y=3.
Find the value of x.
The illustration shows that points A and B are equidistant from point L. By the converse of the Angle Bisector Theorem, we know that L must lie on the angle bisector of ?AYB. This means that ?AYL=?BYL, so we can solve for x as shown below:
So, our answer is x=4.
QS is the angle bisector of ?PQR. Find the value of x.
From the information we've been given, we know that ?PQS is congruent to ?SQR because QS bisects the whole angle, ?PQR. We have been given the measure of the whole angle and the measure of ?SQR, which is half of the entire angle (since the angle has been bisected). Therefore, we have
Thus, we get x=9. If we were to plug in 9 for x, we would see that, indeed, ?SQR is half of ?PQR.
Unformatted text preview: Answers: Chapter 5 Properties and Attributes of Triangles Lesson 5—4 The Triangle Midsegment Theorem; 3. 5.1 4. 11.2 5. 5.6 6. 29° 7. 29° 8. 151° 11. 38 12. 13.5 13. 19 14. 55° 15. 55° 16. 125° 18. 34 19. 17 20. The perimeter of aGHJ is twice the perimeter of aKLM. 21. n = 36 22. n = 26.5 23. n = 8 24. n = 21 25. n = 4 26. n = 5 30. 16.5 31. 11 32. 4.125 33. 57° 34. 57° 35. 123° Answers: Chapter 5 Properties and Attributes of Triangles Lesson 54 The Triangle Midsegment Theorem; 53. 6 54. 8.25 55. 9 Solutions: Chapter 5 Properties and Attributes of Triangles Lesson 5—4 The Triangle Midsegment Theorem; 3. 11. 12. 13. 14. 15. 16. 1B. 19. 20. .NZ NM = lxv 2 1 — 10.2 =5.1 2< ) .XZ:2LM = 2(5.6)=11.2 1 —XZ g =— 11.2 :56 2( ) mALMN = mAMNZ= 29° . m4 YXZ: mAMNZ= 29° . mAXLM+ mALMN = 180 mAXLM + 29 : 180 mAXLM = 151° GJ = 2P0 = 2(19) = 38 mAPQFx’ = mAQFa’J = 55° mAHGJ = mAOFi'J = 55° mi GPO + mAHGJ = 180° mAGPQ + 55 = 180° méGPQ = 125° P=GH+ HJ+ GJ =12+2LJ+2KL :12+2(4)+2(7) =34 P=KL+LM+KM 1 1 =7 —GH —HJ +2 +2 = 7 + 1(12) + l(8) =1? 2 2 The perimeter of A GHJ is twice the perimeter of AKLM. Solutions: Chapter 5 Properties and Attributes of Triangles Lesson 5—4 The Triangle Midsegment Theorem; 21. SH: 2(54) 3!}: 108 n = 36 22. 2(n— 9) = 35 2.3— 18 = 35 2n = 53 = 26.5 23. 2(4n+5) =74 8n+10=74 8n=64 n=8 24. Zn — 28 = 2(9.5) 2n— 23 =19 2n: 42 n: 21 25. 6n=2(n+8) 6n:2n+16 4n=16 n=4 26. 2(5n)=8n+ 10 10n=8n+10 2n=10 n=5 0 C) II |_|. 3: EU 30. ml—sm (33) = 16.5 31. m I II U C") |\J|—L|'\JI—L 0 0:1 (22): 11 l\)|—L 32. E”. || (.3 I (13.5) = 4.125 II MAM—mm; C') c) 33. mZDCG = mACBA = 57° 34. mLGHE= mLHCD mAABC 57° Solutions: Chapter 5 Properties and Attributes of Triangles Lesson 5—4 The Triangle Midsegment Theorem; 35. mAFJG + mAGHE= 180 mAFJG + 57 = 180 mAFJG = 123° 53. NX= 2X8 = 2(3) = 6 54. MR = git/1X _ § _ _ 2(55) _ 8.25 55. NP = 2M? = 2(4.5) = 9 ...
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