Geometry Section 5-1 1112

Chapter 5Relationships in Triangles

Tuesday, February 28, 2012

SECTION 5-1Bisectors of Triangles


Essential Questions

How do you identify and use perpendicular bisectors in triangles?

How do you identify and use angle bisectors in triangles?


Vocabulary1. Perpendicular Bisector:

2. Concurrent Lines:

3. Point of Concurrency:

4. Circumcenter:

5. Incenter:


Vocabulary1. Perpendicular Bisector: A segment that not only cuts

another segment in half, but it also forms a 90° angle at the intersection

2. Concurrent Lines:


4. Circumcenter:

5. Incenter:




2. Concurrent Lines: Three or more lines that intersect at the same point


4. Circumcenter:

5. Incenter:





3. Point of Concurrency: The common point where three or more lines intersect

4. Circumcenter:

5. Incenter:






4. Circumcenter: The concurrent point where the perpendicular bisectors of the sides of a triangle meet

5. Incenter:






4. Circumcenter: The concurrent point where the perpendicular bisectors of the sides of a triangle meet

5. Incenter: The concurrent point where the angle bisectors of the angles of a triangle meet


5.1 - Perpendicular Bisector Theorem

If a point lies on the perpendicular bisector of a segment, the it is equidistant from the endpoints of the segment


5.1 - Perpendicular Bisector Theorem

If a point lies on the perpendicular bisector of a segment, the it is equidistant from the endpoints of the segment

AC = BC


5.2 - Converse of the Perpendicular Bisector Theorem

If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment


5.2 - Converse of the Perpendicular Bisector Theorem

If a point is equidistant from the endpoints of a segment, then it is on the perpendicular bisector of the segment

If WX = WZ, then XY = ZY


5.3 - Circumcenter Theorem

The circumcenter (concurrent point where perpendicular bisectors intersect) is equidistant from the vertices of a

triangle


5.3 - Circumcenter Theorem

The circumcenter (concurrent point where perpendicular bisectors intersect) is equidistant from the vertices of a

triangle

If G is the circumcenter, then GA = GB = GC


5.4 - Angle Bisector Theorem

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


5.4 - Angle Bisector Theorem

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

If AD bisects ∠BAC, BD ⊥ AB, and CD ⊥ AC, then BD = CD


5.5 - Converse of the Angle Bisector Theorem

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


5.5 - Converse of the Angle Bisector Theorem

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

If BD ⊥ AB, CD ⊥ AC, and BD = CD, then AD bisects ∠BAC


5.6 - Incenter Theorem

The incenter (concurrent point where angle bisectors meet) is equidistant from each side of the triangle


5.6 - Incenter Theorem

The incenter (concurrent point where angle bisectors meet) is equidistant from each side of the triangle

If S is the incenter of ∆MNP, then RS = TS = US


Example 1Find each measure.

a. BC b. XY



a. BC

BC = 8.5

b. XY



a. BC

BC = 8.5

b. XY

XY = 6


Example 1

c. PQ

Find each measure.


Example 1

c. PQ

Find each measure.

3x + 1 = 5x − 3


Example 1

c. PQ

Find each measure.

3x + 1 = 5x − 3-3x -3x


Example 1

c. PQ

Find each measure.

3x + 1 = 5x − 3-3x -3x +3+3


Example 1

c. PQ

Find each measure.

3x + 1 = 5x − 3-3x -3x +3+3

4 = 2x


Example 1

c. PQ

Find each measure.

3x + 1 = 5x − 3-3x -3x +3+3

4 = 2xx = 2


Example 1

c. PQ

Find each measure.

3x + 1 = 5x − 3-3x -3x +3+3

4 = 2xx = 2

PQ = 3x + 1


Example 1

c. PQ

Find each measure.

3x + 1 = 5x − 3-3x -3x +3+3

4 = 2xx = 2

PQ = 3x + 1PQ = 3(2) + 1


Example 1

c. PQ

Find each measure.

3x + 1 = 5x − 3-3x -3x +3+3

4 = 2xx = 2

PQ = 3x + 1PQ = 3(2) + 1

PQ = 6 + 1


Example 1

c. PQ

Find each measure.

3x + 1 = 5x − 3-3x -3x +3+3

4 = 2xx = 2

PQ = 3x + 1PQ = 3(2) + 1

PQ = 6 + 1PQ = 7


Example 2A triangular shaped garden is shown. Can a fountain be placed at the circumcenter and still be in the garden?


Example 2A triangular shaped garden is shown. Can a fountain be placed at the circumcenter and still be in the garden?

No, it cannot


QuestionIf you have an obtuse triangle, where will the circumcenter be?

If you have an acute triangle, where will the circumcenter be?

If you have an right triangle, where will the circumcenter be?



It will be outside the triangle







It will be inside the triangle






It will be inside the triangle


It will be on the hypotenuse of the triangle



a. DB b. m∠WYZ

m∠WYX = 28°



a. DB

DB = 5

b. m∠WYZ

m∠WYX = 28°



a. DB

DB = 5

b. m∠WYZ

m∠WYZ = 28°

m∠WYX = 28°


Example 3

c. QS

Find each measure.


Example 3

c. QS

Find each measure.

4x - 1 = 3x + 2


Example 3

c. QS

Find each measure.

4x - 1 = 3x + 2-3x -3x


Example 3

c. QS

Find each measure.

4x - 1 = 3x + 2-3x -3x +1+1


Example 3

c. QS

Find each measure.

4x - 1 = 3x + 2-3x -3x +1+1

x = 3


Example 3

c. QS

Find each measure.

4x - 1 = 3x + 2-3x -3x +1+1

x = 3

QS = 4x - 1


Example 3

c. QS

Find each measure.

4x - 1 = 3x + 2-3x -3x +1+1

x = 3

QS = 4x - 1QS = 4(3) - 1


Example 3

c. QS

Find each measure.

4x - 1 = 3x + 2-3x -3x +1+1

x = 3

QS = 4x - 1QS = 4(3) - 1

QS = 12 - 1


Example 3

c. QS

Find each measure.

4x - 1 = 3x + 2-3x -3x +1+1

x = 3

QS = 4x - 1QS = 4(3) - 1

QS = 12 - 1QS = 11


Example 4Find each measure if S is the incenter of ∆MNP.

a. SU



a. SU SU is a leg in a right triangle




a2 + b2 = c2




a2 + b2 = c2

a2 + 82 = 102




a2 + b2 = c2

a2 + 82 = 102

a2 + 64 = 100




a2 + b2 = c2

a2 + 82 = 102

a2 + 64 = 100

a2 = 36




a2 + b2 = c2

a2 + 82 = 102

a2 + 64 = 100

a2 = 36

a = 6




a2 + b2 = c2

a2 + 82 = 102

a2 + 64 = 100

a2 = 36

a = 6SU = 6



b. m∠SPU



b. m∠SPU An incenter is created at the concurrent point of the angle bisectors




m∠MNP = 28 + 28 = 56°




m∠MNP = 28 + 28 = 56°

m∠NMP = 31+ 31 = 62°




m∠MNP = 28 + 28 = 56°

m∠NMP = 31+ 31 = 62°

m∠MPN = 180− 62 − 56 = 62°




m∠MNP = 28 + 28 = 56°

m∠NMP = 31+ 31 = 62°

m∠MPN = 180− 62 − 56 = 62°

m∠SPU =

12

(62) = 31°




m∠MNP = 28 + 28 = 56°

m∠NMP = 31+ 31 = 62°

m∠MPN = 180− 62 − 56 = 62°

m∠SPU =

12

(62) = 31°

Check: 28 + 28 + 31 + 31 + 31 + 31 = 180Tuesday, February 28, 2012

Check Your Understading

Make sure to review p. 327 #1-8


Problem Set


Problem Set

p. 327 #9-29 odd, 48

"Great opportunities to help others seldom come, but small ones surround us every day." - Sally Koch


Geometry Section 5-1 1112

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Transcript of Geometry Section 5-1 1112