Holt McDougal Geometry
Perpendicular and Angle Bisectors Bisectors & Points of Concurrency
Holt Geometry
Warm Up
Lesson Presentation
Lesson Quiz
Holt McDougal Geometry
Unit 1 Day 26
Holt McDougal Geometry
Perpendicular and Angle Bisectors
Warm Up Construct each of the following.
1. A perpendicular bisector.
2. An angle bisector.
3. Find the midpoint and slope of the segment
(2, 8) and (–4, 6).
Holt McDougal Geometry
Perpendicular and Angle Bisectors
When a point is the same distance from two or more objects, the point is said to be equidistant from the objects.
Holt McDougal Geometry
Perpendicular and Angle Bisectors
Holt McDougal Geometry
Perpendicular and Angle Bisectors
Example 1A: Applying the Perpendicular Bisector
Theorem and Its Converse
Find each measure.
MN
MN = LN
MN = 2.6
Bisector Thm.
Substitute 2.6 for LN.
Holt McDougal Geometry
Perpendicular and Angle Bisectors
Example 1B: Applying the Perpendicular Bisector
Theorem and Its Converse
Find each measure.
BC
Since AB = AC and , is the perpendicular bisector of by the Converse of the Perpendicular Bisector Theorem.
BC = 2CD
BC = 2(12) = 24
Def. of seg. bisector.
Substitute 12 for CD.
Holt McDougal Geometry
Perpendicular and Angle Bisectors
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
Subtract 3x from both sides.
Add 17 to both sides.
Divide both sides by 4.
Substitute the given values.
Holt McDougal Geometry
Perpendicular and Angle Bisectors
Remember that the distance between a point and a line is the length of the perpendicular segment from the point to the line.
Holt McDougal Geometry
Perpendicular and Angle Bisectors
Holt McDougal Geometry
Perpendicular and Angle Bisectors
Example 2A: Applying the Angle Bisector Theorem
Find the measure.
BC
BC = DC
BC = 7.2
Bisector Thm.
Substitute 7.2 for DC.
Holt McDougal Geometry
Perpendicular and Angle Bisectors
Example 2B: Applying the Angle Bisector Theorem
Find the measure.
mEFH, given that mEFG = 50°.
Since EH = GH,
and , bisects
EFG by the Converse
of the Angle Bisector Theorem.
Def. of bisector
Substitute 50° for mEFG.
Holt McDougal Geometry
Perpendicular and Angle Bisectors
Example 2C: Applying the Angle Bisector Theorem
Find mMKL.
, bisects JKL
Since, JM = LM, and
by the Converse of the Angle
Bisector Theorem.
mMKL = mJKM
3a + 20 = 2a + 26
a + 20 = 26
a = 6
Def. of bisector
Substitute the given values.
Subtract 2a from both sides.
Subtract 20 from both sides.
So mMKL = [2(6) + 26]° = 38°
Holt McDougal Geometry
Perpendicular and Angle Bisectors
When three or more lines intersect at one point, the lines are said to be concurrent. The point of concurrency is the point where they intersect.
Holt McDougal Geometry
Perpendicular and Angle Bisectors
A circle that contains all the vertices of a polygon is circumscribed about the polygon.
Holt McDougal Geometry
Perpendicular and Angle Bisectors
A circle inscribed in a polygon intersects each line that contains a side of the polygon at exactly one point.
Holt McDougal Geometry
Perpendicular and Angle Bisectors
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.
Holt McDougal Geometry
Perpendicular and Angle Bisectors
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.
Holt McDougal Geometry
Perpendicular and Angle Bisectors
Example 1A: Using the Centroid to Find Segment
Lengths
In ∆LMN, RL = 21 and SQ =4. Find LS.
LS = 14
Centroid Thm.
Substitute 21 for RL.
Simplify.
Holt McDougal Geometry
Perpendicular and Angle Bisectors
Example 1B: Using the Centroid to Find Segment
Lengths
In ∆LMN, RL = 21 and SQ =4. Find NQ.
Centroid Thm.
NS + SQ = NQ Seg. Add. Post.
12 = NQ
Substitute 4 for SQ.
Multiply both sides by 3.
Substitute NQ for NS.
Subtract from both sides.
Holt McDougal Geometry
Perpendicular and Angle Bisectors
Check It Out! Example 1a
In ∆JKL, ZW = 7, and LX = 8.1. Find KW.
Centroid Thm.
Substitute 7 for ZW.
Multiply both sides by 3. KW = 21
Holt McDougal Geometry
Perpendicular and Angle Bisectors
Check It Out! Example 1b
In ∆JKL, ZW = 7, and LX = 8.1. Find LZ.
Centroid Thm.
Substitute 8.1 for LX.
Simplify. LZ = 5.4
Holt McDougal Geometry
Perpendicular and Angle Bisectors
Example 2: Problem-Solving Application
A sculptor is shaping a triangular piece of iron that will balance on the point of a cone. At what coordinates will the triangular region balance?
Holt McDougal Geometry
Perpendicular and Angle Bisectors
Example 2 Continued
1 Understand the Problem
The answer will be the coordinates of the centroid of the triangle. The important information is the location of the vertices, A(6, 6), B(10, 7), and C(8, 2).
2 Make a Plan
The centroid of the triangle is the point of intersection of the three medians. So write the equations for two medians and find their point of intersection.
Holt McDougal Geometry
Perpendicular and Angle Bisectors
Solve 3
Let M be the midpoint of AB and N be the midpoint of AC.
CM is vertical. Its equation is x = 8. BN has slope 1. Its equation is y = x – 3. The coordinates of the centroid are D(8, 5).
Example 2 Continued
Holt McDougal Geometry
Perpendicular and Angle Bisectors
Look Back 4
Let L be the midpoint of BC. The equation for AL
is , which intersects x = 8 at D(8, 5).
Example 2 Continued
Holt McDougal Geometry
Perpendicular and Angle Bisectors
Check It Out! Example 2
Find the average of the x-coordinates and the average of the y-coordinates of the vertices of ∆PQR. Make a conjecture about the centroid of a triangle.
Holt McDougal Geometry
Perpendicular and Angle Bisectors
Check It Out! Example 2 Continued
The x-coordinates are 0, 6 and 3. The average is 3.
The y-coordinates are 8, 4 and 0. The average is 4.
The x-coordinate of the centroid is the average of the x-coordinates of the vertices of the ∆, and the y-coordinate of the centroid is the average of the y-coordinates of the vertices of the ∆.
Holt McDougal Geometry
Perpendicular and Angle Bisectors
Assignment
Page 180 #2-7
Page 193-195 #1, 3-7, 21-26