Download 5-4: Proofs Using Transitivity

5-4: Proofs Using Transitivity
Geometry (CP)
11/20/06
EX 1: Use a compass and straightedge to construct an equilateral triangle:
C
B
A
D
(1)
(2)
(3)
(4)
Put your compass center on A and your pencil in the hole at B. Draw circle A.
Now put your compass center on B and your pencil on the hole at A. Draw circle B.
Mark the places where the two circles meet points C and D.
Connect segments AB and AC to form ABC .
EX 2: Write a proof (using two-column form) that the triangle you constructed is
equilateral:
Argument:
Conclusions
Justifications
(1) AC = AB
definition of circle (circle A)
(2) AB = BC
Definition of circle (circle B)
(3) AC = BC
Transitive Property of Equality
(4)
Definition of Equilateral Triangle
ABC is equilateral
EX 3: Suppose AB is the bisector of CAD , and AC is the bisector of EAB . Draw a
picture of this scenario and prove mEAC = mBAD .
C
E
Argument:
B
A
D
Conclusions
Justifications
(1) m  CAB = m  BAD
Definition of Angle Bisector
(2) m  EAC = m  CAB
Definition of Angle Bisector
(3) m  BAD = m  EAC
Transitive Property of Equality
Alternate Interior Angles:
t
Interior Angles: 3 , 4 , 5 , 6 .
Exterior Angles: 1 , 2 , 7 , 8 .
m
1
4
n
5
8
2
3
6
Alternate Interior Angles:
4 and 6
3 and 5
7
// Lines => AIA  Theorem:
If two parallel lines are cut by a transversal, then alternate interior angles are congruent.
EX 4: Given: m // n with angles as numbered in the diagram above.
Prove: 3  5
Argument:
Conclusions
(0) m // n
Justifications
Given
(1) 5  1
Parallel Lines => corresponding  
(2) 1  3
Vertical Angles Theorem
(3) 5  3
Transitive Property of Congruence
The theorem also works the other way:
AIA  // Lines Theorem:
If two angles are cut by a transversal and form congruent alternate interior angles, then
the lines are parallel.
EX. 5: Assume measure of angle 5 is 65 degrees. Find the measure of all angles. Name
all sets of alternate interior/alternate exterior angles.
m  5 = 65
m  6 = 65
m  3 = 65
m  2 = 65
Alternate Interior Angles:
 2 and  6
 1 and  8
m  4 = 115
m  1 = 115
Alternate Exterior Angles:
 5 and  3
 4 and  7
EX 5: Given: WHY  TOC , TOC  PAL
Prove: WH  PA
Proof:
Conclusions
Justifications
(1) WHY  TOC , TOC  PAL
(1) GIVEN
(2) WHY  PAL
(2) Trans. Prop. of Congruence
(3) WH  PA
(3) CPCF
EX 6: Given: QN bisects MQO ,
QO bisects NQP
Prove: MQN  OQP
Proof:
Conclusions
(1) QN bisects MQO ,
Justifications
(1) GIVEN
QO bisects NQP
(2) MQN  NQO
(2) Def. of angle bisector
(3) NQO  OQP
(3) Def. of angle bisector
(4) MQN  OQP
(4) Trans. Prop. Of Congruence
Homework: Lesson Master 5-4A: #3-7, 11