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### Unit 6

Properties and Attributes of Triangles

5.1 Perpendicular and Angle Bisectors

Essential Question

Compare and Contrast Perpendicular Bisectors and Angle bisectors.

Example 1A: Applying the Perpendicular Bisector Theorem and Its Converse

Find each measure.

MN

MN = LN

Bisector Thm.

MN = 2.6

Substitute 2.6 for LN.

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

Definition of Perpendicular bisector.

BC = 2(12) = 24

Substitute 12 for CD.

Example 1C: Applying the Perpendicular Bisector Theorem and Its Converse

Find each measure.

TU

TU = UV

Bisector Thm.

3x + 9 = 7x – 17

Substitute the given values.

9 = 4x – 17

Subtract 3x from both sides.

26 = 4x

Add 17 to both sides.

6.5 = x

Divide both sides by 4.

So TU = 3(6.5) + 9 = 28.5.

The perpendicular bisector of a side of a triangle does not always pass through the opposite vertex.

HINTPerpendicular Bisectors intersect at the Circumcenter

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. In the construction, you saw that the three perpendicular bisectors of a triangle are concurrent. This point of concurrency is the circumcenter of the triangle.

The circumcenter can be inside the triangle, outside the triangle, or on the triangle.

The circumcenter of ΔABC is the center of its circumscribed circle. A circle that contains all the vertices of a polygon is circumscribed about the polygon.

DG, EG, and FG are the perpendicular bisectors of ∆ABC. Find GC.

Example 1: Using Properties of Perpendicular BisectorsG is the circumcenter of ∆ABC. By the Circumcenter Theorem, G is equidistant from the vertices of

∆ABC.

GC = CB

Circumcenter Thm.

Substitute 13.4 for GB.

GC = 13.4

MZ is a perpendicular bisector of ∆GHJ.

Check It Out! Example 1aUse the diagram. Find GM.

GM = MJ

Circumcenter Thm.

Substitute 14.5 for MJ.

GM = 14.5

KZ is a perpendicular bisector of ∆GHJ.

Check It Out! Example 1bUse the diagram. Find GK.

GK = KH

Circumcenter Thm.

Substitute 18.6 for KH.

GK = 18.6

Check It Out! Example 1c

Use the diagram. Find JZ.

Z is the circumcenter of ∆GHJ. By the Circumcenter Theorem, Z is equidistant from the vertices of

∆GHJ.

JZ = GZ

Circumcenter Thm.

Substitute 19.9 for GZ.

JZ = 19.9

Based on these theorems, an angle bisector can be defined as the locus of all points in the interior of the angle that are equidistant from the sides of the angle.

Example 2A: Applying the Angle Bisector Theorem

Find the measure.

BC

BC = DC

Bisector Thm.

BC = 7.2

Substitute 7.2 for DC.

and , bisects

EFGby the Converse

of the Angle Bisector Theorem.

Example 2B: Applying the Angle Bisector TheoremFind the measure.

mEFH, given that mEFG = 50°.

Def. of bisector

Substitute 50° for mEFG.

, bisects JKL

by the Converse of the Angle

Bisector Theorem.

Example 2C: Applying the Angle Bisector TheoremFind mMKL.

mMKL = mJKM

Def. of bisector

3a + 20 = 2a + 26

Substitute the given values.

a + 20 = 26

Subtract 2a from both sides.

a= 6

Subtract 20 from both sides.

So mMKL = [2(6) + 26]° = 38°

A triangle has three angles, so it has three angle bisectors. The angle bisectors of a triangle are also concurrent. This point of concurrency is the incenter of the triangle.

The incenter is the center of the triangle’s inscribed circle. A circle inscribedin a polygon intersects each line that contains a side of the polygon at exactly one point.

MP and LP are angle bisectors of ∆LMN. Find the distance from P to MN.

The distance from P to LM is 5. So the distance from P to MN is also 5.

Example 3A: Using Properties of Angle BisectorsP is the incenter of ∆LMN. By the Incenter Theorem, P is equidistant from the sides of ∆LMN.

PM is the bisector of LMN.

Example 3B: Using Properties of Angle BisectorsMP and LP are angle bisectors of ∆LMN. Find mPMN.

mMLN = 2mPLN

mMLN = 2(50°)= 100°

Substitute 50° for mPLN.

mMLN + mLNM + mLMN = 180°

ΔSum Thm.

100+ 20 + mLMN = 180

Substitute the given values.

Subtract 120° from both sides.

mLMN= 60°

Substitute 60° for mLMN.

5.1 Perpendicular and Angle Bisectors

Summary

Compare and Contrast Perpendicular Bisectors and Angle bisectors.

The perpendicular bisector____________.

Perpendicular bisectors intersect at ________ point of concurrency.

The angle bisector____________.

Angle bisectors intersect at ________ point of concurrency.

Both the perpendicular bisector and angle bisector ____________________________.

Medianand Altitude of a Triangle

How can I find the centroid and orthocenter?

Median, Altitude and Midsegment

I can find the centroid by____________.

I can find the orthocenter by _______________.

Midsegment of a Triangle

How does 2/3 relate to Midsegment of a triangle?

Midsegment of a Triangle

2/3 of a ___________ represents the ______ of a Triangle.

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