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# Section 4.3 & 4.4: Proving ?s are Congruent - PowerPoint PPT Presentation

Section 4.3 & 4.4: Proving s are Congruent. Goals. Identify  figures and corresponding parts Prove that 2  are . Anchors. Identify and/or use properties of congruent and similar polygons Identify and/or use properties of triangles. M. Q. N. R. P. S.

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### Section 4.3 & 4.4: Proving s are Congruent

Goals

• Identify  figures and corresponding parts

• Prove that 2  are 

Anchors

• Identify and/or use properties of congruent and similar polygons

• Identify and/or use properties of triangles

Q

N

R

P

S

Side-Side-Side (SSS)  Postulate

• If 3 sides of 1  are  to 3 sides of a 2nd , then the 2 ’s are .

If Side MN  QR

Side NP  RS and

Side PM  SQ

Then MNP QRS

Then we can say: M Q, N  R , and P  S

Reasons

Given: W is the midpoint of QS PQ  TS and PW  TWProve: PQW  TSW

• W is the mdpt of QS,

• PQ  TS and PW  TW

• Given

2) QW  SW

2) Def. of midpoint

3) PQW  TSW

3) SSS

Reasons

Given: D is the midpoint of ACABC is isosceles ABC is the vertex angleProve: ABD  CBD

• D is the mdpt of AC,

• ABC is isosceles

• Given

2) Def. of midpoint

3) AB  BC

3) Property of Isosceles 

4) BD  BD

4) Reflexive

5) ABD  CBD

5) SSS

X

)

P

W

S

Y

)

Side-Angle-Side (SAS)  Postulate

• If 2 sides and the included  of 1  are  to 2 sides and the included  of a 2nd , then the 2 s are .

If Side PQ  WX

Angle Q  X

Side QS  XY

Then PQS WXY

Then we can say: PS  WY, P  W , and S  Y

Reasons

Given: QRS is isosceles RT bisects QRS QRS is the vertex angle Prove: QRT  SRT

)

• QRS is isosceles

• RT bisects QRS

• Given

2) QRT  SRT

2)  bisector

3) QR  RS

3) Property of Isosceles 

4) RT  RT

4) Reflexive

5) QRT  SRT

5) SAS

Reasons

Given: BD and AE bisect each otherProve: ABC  EDC

)

)

• BD and AE bisect

• each other

• Given

2) BC  CD, AC  CE

2) Segment bisectors

3) BCA  ECD

3) Vertical angles

4) ABC  EDC

4) SAS

Q

M

)

R

N

S

P

)

)

Angle-Side-Angle (ASA)  Postulate

• If 2 ’s and the included side of 1  are  to 2 ’s and the included side of a 2nd, then the 2  are 

If Angle  N   R

Side MN  QR

Angle  M   Q

Then MNP QRS

Then we can say: MP  QS, NP  RS , and P  S

)

Statements

Reasons

Given: B  N RW bisects BNProve: BRO  NWO

)

)

• B  N

• RW bisects BN

• Given

2) BOR  WON

2) Vertical Angles

3) BO  ON

3) Segment bisector

4) BRO  NWO

4) ASA

1

3

4

2

Statements

Reasons

Given: 1  2 CD bisects BCEProve: BCD  ECD

)

)

• 1  2

• CD bisects BCE

• Given

2) 3  4

• Supplements of congruent s are congruent

3) BCD  ECD

3) Angle bisector

4) BCE is isosceles

4) Property of isosceles 

5) BC  CE

5) Property of isosceles 

6) BCD  ECD

6) ASA

Q

(

(

W

P

Y

S

(

(

Angle-Angle-Side (AAS)  Theorem

• If 2 ’s and a non-included sideof 1  are  to 2  ‘s and a non-included side of a 2nd , then the 2 ’s are .

If Angle P  W

Angle S  Y

Side QP  WX

Then PQS WXY

Then we can say: QS  XY, PS  WY , and Q  X

)

Statements

Reasons

Given: AD ║ EC , B is the mdpt of CDProve: ABD  EBC

)

)

1) AD ║ EC , B is the mdpt of CD

• Given

2) A  E

2) Alternate Interior s

3) ABD  CBE

3) Vertical Angles

4) BD  BC

4) Midpoint

5) ABD  EBC

5) AAS

)

Statements

Reasons

Given: AD ║ EC , B is the mdpt of CDProve: ABD  EBC

)

)

1) AD ║ EC , B is the mdpt of CD

• Given

2) A  E, D  C

2) Alternate Interior s

3) BD  BC

3) Midpoint

4) ABD  EBC

4) AAS

40

40

50

50

Why Angle-Angle-Angle (AAA)Doesn’t Work

The angles are , but the sides are proportional.

(

A

D

F

(

C

B

Why Side-Side-Angle (SSA)Doesn’t Work

Two different triangles can be formed if you use two sides and a non-included angle.

Theorem 4.8: Hypotenuse-Leg (HL)  Theorem

• If the hypotenuse and a leg of a right  are  to a hypotenuse and a leg of a 2nd right , then the 2 ’s are 

D

A

If BC  EF and AC  DF,

then ABC  DEF

Special case of SSA

B

C

E

F

Then we can say: AB  DE, A  D , and C  F

Reasons

Given: RS  QT QRT is isosceles QRT is the vertex angleProve: QRS  TRS

1) RS  QT, QRT is isosceles

• Given

2) QSR  90,

TSR  90

2) Definition of perpendicular

3) QSR  TSR

3) Substitution

4) QR  RT

4) Property of isosceles 

5) RS  RS

5) Reflexive

6) QRS  TRS

6) HL