Bisectors in Triangles

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RT || HI , RS || GI , S T || HG. Bisectors in Triangles. Lesson 5-2. Lesson 5-1 Quiz – Midsegments of Triangles. In GHI , R , S , and T are midpoints. 1. Name all the pairs of parallel segments. 2. If GH = 20 and HI = 18, find RT .

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RT || HI, RS || GI, ST || HG

Bisectors in Triangles

Lesson 5-2

Lesson 5-1 Quiz – Midsegments of Triangles

In GHI, R, S, and T are midpoints.

1. Name all the pairs of parallel segments.

2. If GH = 20 and HI = 18, find RT.

3. If RH = 7 and RS = 5, find ST.

4. If m G = 60 and m I = 70, find m GTR.

5. If m H = 50 and m I = 66, find m ITS.

6. If m G = m H = m I and RT = 15, find the perimeter of GHI.

9

7

70

64

90

5-1

Bisectors in Triangles

Lesson 5-2

When a point is the same distance from two or more

objects, the point is said to be equidistantfrom

the objects.

Triangle congruence theorems can be

used to prove theorems about equidistant points.

5-2

Bisectors in Triangles

Lesson 5-2

The shortest segment from a point to a line is perpendicular to the line. This fact is used to define the distance from a point to a lineas the length of the perpendicular segment from the point to the line.

5-2

Use the map of Washington, D.C. Describe the set of points that are equidistant from the Lincoln Memorial and the Capitol.

Bisectors in Triangles

Lesson 5-2

Real-World Connection

The Converse of the Perpendicular Bisector Theorem states If a point

is equidistant from the endpoints of a segment, then it is on the

perpendicular bisector of the segment.

Quick Check

5-2

(continued)

A point that is equidistant from the Lincoln Memorial and the Capitol

must be on the perpendicular bisector of the segment whose

endpoints are the Lincoln Memorial and the Capitol.

Bisectors in Triangles

Lesson 5-2

Therefore, all points on the perpendicular bisector of the segment

whose endpoints are the Lincoln Memorial and the Capitol are

equidistant from the Lincoln Memorial and the Capitol.

Quick Check

5-2

Find x, FB, and FD in the diagram above.

Bisectors in Triangles

Lesson 5-2

Using the Angle Bisector Theorem

FD = FB Angle Bisector Theorem

7x – 37 = 2x + 5Substitute.

7x = 2x + 42 Add 37 to each side.

5x = 42 Subtract 2x from each side.

x = 8.4 Divide each side by 5.

FB = 2(8.4) + 5 = 21.8 Substitute.

FD = 7(8.4) – 37 = 21.8 Substitute.

Quick Check

5-2

No; if CN = DN, CNBDNB by SAS and CB = DB by CPCTC, which

is false.

Bisectors in Triangles

Lesson 5-2

Lesson Quiz

Use this figure for Exercises 1–3.

1. Find BD.

2. Complete the statement:

C is equidistant from ? .

3. Can you conclude that

CN = DN? Explain.

Use this figure for Exercises 4–6.

4. Find the value of x.

5. Find CG.

6. Find the perimeter of

16

6

8

points A and B

48

5-2

Q P.

Bisectors in Triangles

Lesson 5-2

Check Skills You’ll Need

(For help, go to Lesson 1-7.)

1. Draw a triangle, XYZ. Construct STV so that STVXYZ.

2. Draw acute P. Construct Q so that

3. Draw AB. Construct a line CD so that CDAB and CD bisects AB.

4. Draw acute angle E. Construct the bisector of E.

TM bisects STU so that mSTM = 5x + 4 and mMTU = 6x – 2.

5. Find the value of x.

6. Find mSTU.

Use a compass and a straightedge for the following.

Check Skills You’ll Need

5-2

Bisectors in Triangles

Lesson 5-2

Check Skills You’ll Need

Solutions

1-4. Answers may vary. Samples given:

1. 2.

3. 4.

5. Since TM bisects STU, mSTM = mMTU. So, 5x + 4 = 6x – 2.

Subtract 5x from both sides: 4 = x – 2; add 2 to both sides: x = 6.

6. From Exercise 5, x = 6. mSTU = mSTM + mMTU = 5x + 4 + 6x – 2

= 11x + 2 = 11(6) + 2 = 68.

5-2