Postulate Addition Segnent 2 Postulate POSTULATE: A C, is … · Distance = |a – b| SEGMENT...

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POSTULATE: A rule this is accepted without proof because it’ s something obvious. PARALLEL LINE POSTULATE: Through a point not a given line, there is a line parallel to the given line. THEOREM: A rule that is proven to be true. PARALLEL LINE POSTULATE: Through a point not a given line, there is a line parallel to the given line RULER POSTULATE 1. Any two points can have coordinates 0 and 1 2. Distance = |a – b| SEGMENT ADDITION POSTULE: If B is in between A and C, then AB + BC = AC CONGRUENT SEGMENTS: Segments that have same shape and equal lengths. CONGRUENT Same Shape and Equal in Size DE = FG DE = 3 FG = 3 MIDPOINT DEFINTION: A point that divides the segment into two congruent segments. IF M is the midpoint of , then and MIDPOINT FORMULA: (A + B) / 2 If B is midpoint in between A and C, and A = -1, C = -5 Then B = [-1 + (-5)] / 2 = (-6)/2 = -3 MIDOINT THEOREM: If M is the midpoint of then _________________________________________________________________________________ DE FG AB AM ! " ### MB !" ## AM = MB AB AM = MB = 1 2 AB

Transcript of Postulate Addition Segnent 2 Postulate POSTULATE: A C, is … · Distance = |a – b| SEGMENT...

Page 1: Postulate Addition Segnent 2 Postulate POSTULATE: A C, is … · Distance = |a – b| SEGMENT ADDITION POSTULE: If B is in between A and C, then AB + BC = AC CONGRUENT SEGMENTS: ...

POSTULATE: A rule this is accepted without proof because it’ s something obvious. PARALLEL LINE POSTULATE: Through a point not a given line, there is a line parallel to the given line. THEOREM: A rule that is proven to be true. PARALLEL LINE POSTULATE: Through a point not a given line, there is a line parallel to

the given line RULER POSTULATE 1. Any two points can have coordinates 0 and 1

2. Distance = |a – b| SEGMENT ADDITION POSTULE: If B is in between A and C, then AB + BC = AC CONGRUENT SEGMENTS: Segments that have same shape and equal lengths.

CONGRUENT Same Shape and Equal in Size

DE = FG DE = 3 FG = 3 MIDPOINT DEFINTION: A point that divides the segment into two congruent segments.

IF M is the midpoint of , then and MIDPOINT FORMULA: (A + B) / 2

If B is midpoint in between A and C, and A = -1, C = -5

Then B = [-1 + (-5)] / 2 = (-6)/2 = -3

MIDOINT THEOREM: If M is the midpoint of then_________________________________________________________________________________

Using a number line involves the following basic assumptions. Statementssuch as these that are accepted without proof are called postulates or axioms.

Postulate 7 Ruler PostulateThe points on a line can be paired with the real numbers in such a way that:

1. Any two desired points can have coordinates 0 and 1.

2. The distance between any two points equals the absolute value of the differ-ence of their coordinates.

Postulate 2 Segnent Addition PostulateIf B is between A rnd C, then

AB+BC-AC.

Example B is between A andAC = 12. Find:a. The value ofx

Solation a. AB + BC = ACx+(xa6):12

2x*6:122x:6

:=3

C, with AB = x, BC = x + 6, and

b. BCb. BC:x*6

-3+6-9

Congruent segments are segments that have equallengths. To indicate that DE and FG are congruent, youcan write eitherDE : FG o, DE - F?(read " DE iscongruent to FZ").Note that these statements are equivalent and may beused interchangeably.

The midpoint of a segment is the point that divides thesegment into two congruent segments. In the figure,AP: PB. P is the midpoint of AB.

A bisector of a segment is a line, segment, ray, or planethat intersects the segment at its midpoint. Line / is abisector of AB. PQ ard plane X also bisect lB.

.2a2.

Points, Lines, Planes, and Angles / 7

DE ≅ FG

AB AM! "###

≅ MB! "##

AM = MB

AB AM = MB = 12AB

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CONSTRUCTIONS www.mathopenref.com CIRCLE: Set compass to a specific radius length. Construct a Circle. The distance from the center of the circle to the circle is the __________. All points on a circle are _____________ from the center CONGRUENT SEGMENTS

CONSTRUCTING SEGMENTS WITH SPECIFIC LENGHTS

Basic ConstructionsObiectirses

1. Perform seven basic constructions.2. Use these basic constructions in original construction exercises.3. State and apply theorems involving concurrent lines.

8-1 What Construction MeansIn Chapters l-7 we have used rulers and protractors to draw segments withcertain lengths and angles with certain measures. In this chapter we will con'struct Eeometric figures using only two instruments, a straightedge and a com-pass. (You may use a ruler as a straightedge as long as you do not use themarks on the ruler.)

Using a Straightedge in ConstructionsGiven two points A and B, we know from Postulate 6 that there isexactly one line through ,4 and B. We agree that we can use a straight-edge to draw iE or parts of the line, such as lS ana AB.

Using a Compass in ConstructionsGiven a point O and a length r, we know from the definition of a circlethat there is exactly one circle with center O and radius r. We agreethat we can use a compass to draw this circle or arcs of the circle.

Construction 7Given a segment, construct a segment congruent to the given segment.

Given: ABConstruct: A segment congruent to ,,q.A

Procedure:L Use a straightedge to draw a line. Call it /.

2. Choose any point on / and label it X.3. Set your compass for radius ,48. Using X as center,

draw an arc intersecting line l. Label the point ofintersection X

XY is congruent to l,B.

Justification: Since AB was used for the radius of OX, X)'is congruentto AB.

Constructions and Loci / 335Written ExercisesOn your paper, draw two segments roughly like those shown. Use these lengthsin Exercises l-4 to construct a segment having the indicated length.

A l.a+b 2.b-a 3.3a-b 4.a+2b5. Using any convenient length for a side, construct an equilateral triangle.6. a. Construct a 30' angle. b. Construct a 15' angle.

7. Draw any acute LACU. Use a method based on the SSS Postulate toconstruct a triangle congruent to LACU.

8. Draw any obtuse LOBT. Use the SSS method to construct a trianglecongruent to LOBT.

9. Repeat Exercise 7, but use the SAS method.10. Repeat Exercise 8, but use the ASA method.

On your paper, draw two angles roughly like those shown. Then for Exercises11-14 construct an angle having the indicated measure.

ll. x *y 12.x-v t3. f;x 14. l8o - 2y

B 15. a. Draw any acute triangle. Bisect each of the three angles.b. Draw any obtuse triangle. Bisect each of the three angles'c. What do you notice about the points of intersection of the bisectors in

parts (a) and (b)?16. Construct a six-pointed star using the following procedure.

l. Draw a ray, AE. On AB mark off, in order, points C and D such thatAB_BC-CD.

2. Construct equilateral'LADG.3. On ,4Ci mark off points ,E and -F such that AE - EF = FG. (Note that

AE = AB.)4. On GD mark off points H and l such that GH - HI = ID.5. To complete the star, draw the three lines fE, Ei, andti.

Construct an angle having the indicated measure.

17. l2o 18. 150 19. 165 20. 45

21. Draw any LABC. Construct LDEF so that LDEF - LABC arrdDE = 2AB.

22. Construct a ARSZ such that RS:S7: TR = 4:6:7.

338 / Chapter 8

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PERPENDICULAR BISECTOR

on your paper draw figures roughly like those shown. Use them in constructingthe ligures described in Exercises 23-25.

-___L____=--,d.

23. An isosceles triangle with a vertex angle of n" and legs d24. An isosceles triangle with a vertex angle of n" arrd base s

c 25. A parallelogram with an n" angle, longer side s, and longer diagonar d* 26. on your paper draw figures roughly like the ones shown. Then construct a

triangle whose three angles are congruent to 11, L2, and /_3, and whosecircumscribed circle has radius r.

8-2 Perpendiculars and ParallelsThe next three constructions are based on the following theorem and postulate.(l) If a point is equidistant from the endpoints of a segment, then the point lies

on the perpendicular bisector of the segment.(2) Through any two points there is exactly one line.

Construction 4Given a segment, construct the perpendicular bisector of the segment

Given: AB

Construct: The perpendicular bisector of AB

Procedure:l. Using any radius greater thanlAB, draw four arcs of

equal radii, two with center I and two with center B.Label the points of intersection of these arcs X and y.

2. Draw 7i.€AXI is the perpendicular bisector of AB.

Justification: Points X and Y are equidistant from,4 andB. Thus i7 is ttre perpendicular bisectorof AB.

Constructions and Loci / 339Construction 5Given a point on a line, construct the perpendicular to the line at the given point.

Given: Point C on line kConstruct: The perpendicular to k at C

Procedure:1. Using C as center and any radius, draw arcs intersect-

ing k at X and Y.

2. Using X as center and a radius greater than CX, drawan arc. Using Y as center and the same radius, drawat arc intersecting the first arc at Z.

3. Draw tZ.E i, p"rp"ndicular to k at C.

Justification: Points X and Y were constructed so that C is equidistant from Xand X Then point Z was constructed so that Z is equidistantfrom X and Y. Sirce tZ is the perpendicular bisector of {y, tZis perpendicular to k at C.

C

Construction 6Given a point outside a line, construct the perpendicular to the line from thegiven point.

Given: Point P outside line ftConstruct: The perpendicular to k from P

Procedure:1. Using P as center, draw two arcs of equal radii that

intersect k at points X and Y.

2. Using X and Y as centers and a suitable radius, drawarcs that intersect at a point Z.

3. Draw FZ.

FZ i" p"rp"ndicular to ft.

Justification: Both P and Z are equidistant from X andX Thus PZ is the pernendicular bisectorof xY, una FZ t i. I

P.

3& / Chapter 8

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REVIEW HW

Chapter 1Foundations for Geometry

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Chapter 1Foundations for Geometry

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