Perpendicular and Angle BisectorsBisectors & Points … · Holt McDougal Geometry Perpendicular and Angle BisectorsBisectors & Points of Concurrency Holt Geometry Warm Up Lesson Presentation

Holt McDougal Geometry

Perpendicular and Angle Bisectors Bisectors & Points of Concurrency

Holt Geometry

Warm Up

Lesson Presentation

Lesson Quiz


Unit 1 Day 26


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).



When a point is the same distance from two or more objects, the point is said to be equidistant from the objects.





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.



Example 1B: Applying the Perpendicular Bisector


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.



Example 1C: Applying the Perpendicular Bisector


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.



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.

Substitute 7.2 for DC.



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.



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°



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.



A circle that contains all the vertices of a polygon is circumscribed about the polygon.



A circle inscribed in a polygon intersects each line that contains a side of the polygon at exactly one point.



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.



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.



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.



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



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



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?



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.



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



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



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.



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 ∆.



Assignment

Page 180 #2-7

Page 193-195 #1, 3-7, 21-26

Documents

Perpendicular and Angle BisectorsBisectors & Points … · Holt McDougal Geometry Perpendicular and Angle BisectorsBisectors & Points of Concurrency Holt Geometry Warm Up Lesson Presentation