Prove and apply theorems about
perpendicular bisectors.
Prove and apply theorems about angle
bisectors.
5.1 Objectives
Example 1A: Applying the Perpendicular Bisector
Theorem and Its Converse
Find each measure.
MN
MN = LN
MN = 2.6
 Bisector Thm.
Substitution
Example 1C: Applying the Perpendicular Bisector
Theorem and Its Converse
TU
Find each measure.
So TU = 3(6.5) + 9 = 28.5.
TU = UV  Bisector Thm.
3x + 9 = 7x – 17
9 = 4x – 17
26 = 4x
6.5 = x
Subtraction POE
Addition POE.
Division POE.
Substitution
Check It Out! Example 1b
Given that DE = 20.8, DG = 36.4,
and EG =36.4, which Theorem
would you use to find EF?
Find the measure.
Since DG = EG and , is the
perpendicular bisector of by
the Converse of the Perpendicular
Bisector Theorem.
Remember that the distance between a point and a
line is the length of the perpendicular segment from
the point to the line.
Example 2A: Applying the Angle Bisector Theorem
Find the measure. BC
BC = DC
BC = 7.2
 Bisector Thm.
Substitution
Find the measure.
mEFH, given that mEFG = 50°.
Since EH = GH,
and , bisects
EFG by the Converse
of the Angle Bisector Theorem.
Example 2C: Applying the Angle Bisector Theorem
Find mMKL.
, bisects JKL
Since, JM = LM, and
by the Converse of the Angle
Bisector Theorem.
mMKL = mJKM
3a + 20 = 2a + 26
a + 20 = 26
a = 6
Def. of  bisector
Substitution.
Subtraction POE
Subtraction POE
So mMKL = [2(6) + 26]° = 38°
Check It Out! Example 2a
Given that mWYZ = 63°, XW = 5.7,
and ZW = 5.7, find mXYZ.
mWYZ = mWYX
mWYZ + mWYX = mXYZ
mWYZ + mWYZ = mXYZ
2(63°) = mXYZ
126° = mXYZ
2mWYZ = mXYZ
Prove and apply properties of
perpendicular bisectors of a triangle.
Prove and apply properties of angle
bisectors of a triangle.
5.2 Objectives
The perpendicular bisector of a side of a triangle
does not always pass through the opposite
vertex.
Helpful Hint
A median of a triangle is a segment whose
endpoints are a vertex of the triangle and the
midpoint of the opposite side.
Every triangle has three medians, and the medians
are concurrent.
The point of concurrency of the medians of a triangle
is the centroid of the triangle . The centroid is
always inside the triangle. The centroid is also called
the center of gravity because it is the point where a
triangular region will balance.
The length of the
segment from the vertex
to the centroid is twice
the length of the
segment from the
centroid to the midpoint
Example 1B: Using the Centroid to Find Segment
Lengths
In ∆LMN, RS = 5
Find SL and RL.
.
SL = 10 and RL = 15
Check It Out! Example 1a
In ∆JKL, ZK = 14,
Find ZW and WK
ZW = 7 and WK = 21
Check It Out! Example 1b
In ∆JKL, JY = 36,
Find JZ and ZY.
JZ = 24 and ZY = 12
Lesson Drill
Use the figure for Items 1–3. In ∆ABC, AE = 12,
DG = 7, and BG = 9. Find each length.
1. AG
2. GC
3. GF
8
14
13.5

Bisector_and_Centroid_of_a_Triangle.ppt

  • 1.
    Prove and applytheorems about perpendicular bisectors. Prove and apply theorems about angle bisectors. 5.1 Objectives
  • 3.
    Example 1A: Applyingthe Perpendicular Bisector Theorem and Its Converse Find each measure. MN MN = LN MN = 2.6  Bisector Thm. Substitution
  • 4.
    Example 1C: Applyingthe Perpendicular Bisector Theorem and Its Converse TU Find each measure. So TU = 3(6.5) + 9 = 28.5. TU = UV  Bisector Thm. 3x + 9 = 7x – 17 9 = 4x – 17 26 = 4x 6.5 = x Subtraction POE Addition POE. Division POE. Substitution
  • 5.
    Check It Out!Example 1b Given that DE = 20.8, DG = 36.4, and EG =36.4, which Theorem would you use to find EF? Find the measure. Since DG = EG and , is the perpendicular bisector of by the Converse of the Perpendicular Bisector Theorem.
  • 6.
    Remember that thedistance between a point and a line is the length of the perpendicular segment from the point to the line.
  • 7.
    Example 2A: Applyingthe Angle Bisector Theorem Find the measure. BC BC = DC BC = 7.2  Bisector Thm. Substitution Find the measure. mEFH, given that mEFG = 50°. Since EH = GH, and , bisects EFG by the Converse of the Angle Bisector Theorem.
  • 8.
    Example 2C: Applyingthe Angle Bisector Theorem Find mMKL. , bisects JKL Since, JM = LM, and by the Converse of the Angle Bisector Theorem. mMKL = mJKM 3a + 20 = 2a + 26 a + 20 = 26 a = 6 Def. of  bisector Substitution. Subtraction POE Subtraction POE So mMKL = [2(6) + 26]° = 38°
  • 9.
    Check It Out!Example 2a Given that mWYZ = 63°, XW = 5.7, and ZW = 5.7, find mXYZ. mWYZ = mWYX mWYZ + mWYX = mXYZ mWYZ + mWYZ = mXYZ 2(63°) = mXYZ 126° = mXYZ 2mWYZ = mXYZ
  • 11.
    Prove and applyproperties of perpendicular bisectors of a triangle. Prove and apply properties of angle bisectors of a triangle. 5.2 Objectives
  • 12.
    The perpendicular bisectorof a side of a triangle does not always pass through the opposite vertex. Helpful Hint
  • 13.
    A median ofa triangle is a segment whose endpoints are a vertex of the triangle and the midpoint of the opposite side. Every triangle has three medians, and the medians are concurrent.
  • 14.
    The point ofconcurrency of the medians of a triangle is the centroid of the triangle . The centroid is always inside the triangle. The centroid is also called the center of gravity because it is the point where a triangular region will balance. The length of the segment from the vertex to the centroid is twice the length of the segment from the centroid to the midpoint
  • 15.
    Example 1B: Usingthe Centroid to Find Segment Lengths In ∆LMN, RS = 5 Find SL and RL. . SL = 10 and RL = 15
  • 16.
    Check It Out!Example 1a In ∆JKL, ZK = 14, Find ZW and WK ZW = 7 and WK = 21
  • 17.
    Check It Out!Example 1b In ∆JKL, JY = 36, Find JZ and ZY. JZ = 24 and ZY = 12
  • 18.
    Lesson Drill Use thefigure for Items 1–3. In ∆ABC, AE = 12, DG = 7, and BG = 9. Find each length. 1. AG 2. GC 3. GF 8 14 13.5