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GEOMETRY Connections 31
PROOF #12#
A proof convinces an audience that a conjecture is true for ALL cases (situations) that fit the conditions of the conjecture. For example, "If a polygon is a triangle on a flat surface, then the sum of the measures of the angles is 180˚." Because we proved this conjecture in chapter two, it is always true. There are many formats that may be used to write a proof. This course explores three of them, namely, paragraph, flow chart, and two-column.
Example
If BDis a perpendicular bisector of AC, prove that ∆ABC isosceles.
Paragraph proof
To prove that ∆ABC is isosceles, show that BA! BC. We can do this by showing that the two segments are corresponding parts of congruent triangles.
Since BD is perpendicular to AC , m∠BDA = m∠BDC = 90°. A
B
D^ C
Since BD bisects AC , AD! CD. With BD! BD (reflexive property), ∆ADB ≅ ∆CDB by SAS.
Finally, BA! BC because corresponding parts of congruent triangles are congruent. Therefore, ∆ABC must be isosceles since two of the three sides are congruent.
Flow chart proof
Given: BD is the perpendicular bisector of AC
AD CD
ADB BDC BD BD
!CBD !ABD
!ABC is isosceles
BA BC
bisector (^) reflexive
SAS !s have parts
!
!!
!
!
" !"
Definition of isosceles
!
Two-Column Proof Given: BDis a bisector of AC. BD is perpendicular to AC. Prove: ∆ ABC is isosceles Statement Reason BD bisects AC. Given BD! AC Given AD! CD Def. of bisector ∠ADB and ∠BDC Def. of perpendicular are right angles ∠ADB ≅ ∠BDC All right angles are ≅. BD! BD Reflexive property ∆ABD ≅ ∆CBD S.A.S. AB! CB ≅ ∆'s have ≅ parts ∴∆ABC is isosceles Def. of isosceles
32 Extra Practice
In each diagram below, are any triangles congruent? If so, prove it. (Note: It is good practice to try different methods for writing your proofs.)
A
B
D^ C
A
B
C
D
E
A
B
C D
A B
C
D 5.
B A
C^ D 6.
A
B
C
D
E
F
Complete a proof for each problem below in the style of your choice.
- Given: TRand MNbisect each other. Prove: !NTP " !MRP
N
T
R
P M
- Given: CDbisects !ACB;! 1 "! 2. Prove: !CDA " !CDB
C
B
D
A
1 2
- Given: AB|| CD, !B " !D, AB! CD Prove: !ABF " !CED
A
F
B
E
C
D
- Given: PG! SG,TP! TS Prove: !TPG " !TSG
P
T
S
G
- Given: OE! MP, OE bisects!MOP
Prove: !MOE " !POE M
O^ E
P
12. Given: AD||BC,DC||BA
Prove: !ADB " !CBD
A B
D^ C
34 Extra Practice
- Yes! BCD! BCA BC BC
! ABC! DBC
Given Reflexive
right! 's are =
SAS
AC "CD
"
"
(^) " 4. Yes
Reflexive
BA CD
! ABC! CDA
Given
SSS
AD BC CA CA^ Given
!
!
! !
- Not necessarily. Counterexample: 6. Yes Given Given AC FD
! ABC! DEF HL
BC EF
!
!!
- NP! MPand TP! RPby definition of bisector. !NPT " !MPRbecause vertical angles are equal. So, !NTP " !MRPby SAS.
- !ACD " !BCDby definition of angle bisector. CD! CD by reflexive so !CDA " !CDBby ASA.
- !A " !Csince alternate interior angles of parallel lines congruent so !ABF " !CED by ASA.
- TG! TG by reflexive so !TPG " !TSGby SSS.
- !MEO " !PEObecause perpendicular lines form !right angles !MOE " !POEby angle
bisector and OE! OEby reflexive. So, !MOE " !POEby ASA.
- !CDB " !ABDand !ADB " !CBDsince parallel lines give congruent alternate interior angles.
DB! DB by reflexive so !ADB " !CBDby ASA.
- DB! EBby definition of bisector. !DBA " !EBCsince vertical angles are congruent. So !ADB " !CEB by AAS.
- !RQP " !SQPsince perpendicular lines form congruent right angles. PQ! PQby reflexive so !PQR " !PQS^ by AAS.
- !SQP " !RQPby angle bisector and PQ! PQby reflexive, so !SPQ " !RPQby AAS.
- !KYT " !HUGbecause parallel lines form congruent alternate exterior angles. TY+YU=YU+GU so TY! GUby subtraction. !T " !G since perpendicular lines form congruent right angles. So !KTY " !HGU by ASA. Therefore, !K " !Hsince !triangles have congruent parts.
- !MQL " !WLQ since parallel lines form congruent alternate interior angles. QL! Q L by reflexive so !MQL " !WLQby SAS so !WQL " !MLQsince congruent triangles have congruent parts. So ML || WQsince congruent alternate interior angles are formed by parallel lines.
- Yes!
Reflexive
! DC DC
BDC ECD
! BCD! EDC SAS
DB "CE " "
"
- Not necessarily.
- Not necessarily.
