Transcript Example 1 - Xavier High School

4-3 Using Congruent Triangles
Using Congruent Triangles
•
Our goal in this section is to deduce information about segments or angles once we have shown that
they are corresponding parts of congruent triangles.
C
Example 1:
Given: AB and CD bisect each other at M.
Prove: AD
A
M
BC
B
D
Statements
Reasons
AB and CD bisect each other at M
M is the midpoint of AB and CD
AM = MB; DM = MC
< AMD = < BMC
AMD =
<A=<B
AD
BC
BMC
Using Congruent Triangles
•
Our goal in this section is to deduce information about segments or angles once we have shown that
they are corresponding parts of congruent triangles.
C
Example 1:
Given: AB and CD bisect each other at M.
Prove: AD
A
M
BC
B
D
Statements
Reasons
AB and CD bisect each other at M
Given
M is the midpoint of AB and CD
AM = MB; DM = MC
< AMD = < BMC
AMD =
<A=<B
AD
BC
BMC
Using Congruent Triangles
•
Our goal in this section is to deduce information about segments or angles once we have shown that
they are corresponding parts of congruent triangles.
C
Example 1:
Given: AB and CD bisect each other at M.
Prove: AD
A
M
BC
B
D
Statements
Reasons
AB and CD bisect each other at M
Given
M is the midpoint of AB and CD
Def. of a bisector of a segment
AM = MB; DM = MC
< AMD = < BMC
AMD =
<A=<B
AD
BC
BMC
Using Congruent Triangles
•
Our goal in this section is to deduce information about segments or angles once we have shown that
they are corresponding parts of congruent triangles.
C
Example 1:
Given: AB and CD bisect each other at M.
Prove: AD
A
M
BC
B
D
Statements
Reasons
AB and CD bisect each other at M
Given
M is the midpoint of AB and CD
Def. of a bisector of a segment
AM = MB; DM = MC
Def. of Midpoint
< AMD = < BMC
AMD =
<A=<B
AD
BC
BMC
Using Congruent Triangles
•
Our goal in this section is to deduce information about segments or angles once we have shown that
they are corresponding parts of congruent triangles.
C
Example 1:
Given: AB and CD bisect each other at M.
Prove: AD
A
M
BC
B
D
Statements
Reasons
AB and CD bisect each other at M
Given
M is the midpoint of AB and CD
Def. of a bisector of a segment
AM = MB; DM = MC
Def. of Midpoint
< AMD = < BMC
Vertical angles congruent
AMD =
<A=<B
AD
BC
BMC
Using Congruent Triangles
•
Our goal in this section is to deduce information about segments or angles once we have shown that
they are corresponding parts of congruent triangles.
C
Example 1:
Given: AB and CD bisect each other at M.
Prove: AD
A
M
BC
B
D
Statements
Reasons
AB and CD bisect each other at M
Given
M is the midpoint of AB and CD
Def. of a bisector of a segment
AM = MB; DM = MC
Def. of Midpoint
< AMD = < BMC
Vertical angles congruent
AMD =
<A=<B
AD
BC
BMC
SAS postulate
Using Congruent Triangles
•
Our goal in this section is to deduce information about segments or angles once we have shown that
they are corresponding parts of congruent triangles.
C
Example 1:
Given: AB and CD bisect each other at M.
Prove: AD
A
M
BC
B
D
Statements
Reasons
AB and CD bisect each other at M
Given
M is the midpoint of AB and CD
Def. of a bisector of a segment
AM = MB; DM = MC
Def. of Midpoint
< AMD = < BMC
Vertical angles congruent
AMD =
<A=<B
AD
BC
BMC
SAS postulate
CPCTC
Using Congruent Triangles
•
Our goal in this section is to deduce information about segments or angles once we have shown that
they are corresponding parts of congruent triangles.
C
Example 1:
Given: AB and CD bisect each other at M.
Prove: AD
A
M
BC
B
D
Statements
Reasons
AB and CD bisect each other at M
Given
M is the midpoint of AB and CD
Def. of a bisector of a segment
AM = MB; DM = MC
Def. of Midpoint
< AMD = < BMC
Vertical angles congruent
AMD =
BMC
SAS postulate
<A=<B
CPCTC
AD
Two lines cut by a transversal and alt. interior
are congruent
BC
Using Congruent Triangles
•
Some proofs require the idea of a line perpendicular to a plane.
A line and a plane are perpendicular if and only if they intersect and the line is perpendicular to
all lines in the plane that pass through the point of intersection
Example 2: Look at book for picture
Statement
PO
PO
Reason
plane x
OA;
PO
OB
< POA = 90; <POB = 90
< POA = < POB
AO =
BO
PO = PO
POA
PA = PB
=
given PO
plane x;
AO = BO
POB
Prove: PA = PB
Using Congruent Triangles
•
Some proofs require the idea of a line perpendicular to a plane.
A line and a plane are perpendicular if and only if they intersect and the line is perpendicular to
all lines in the plane that pass through the point of intersection
Example 2: Look at book for picture
given PO
plane x;
AO = BO
Statement
Reason
PO
Given
PO
plane x
OA;
PO
OB
< POA = 90; <POB = 90
< POA = < POB
AO =
BO
PO = PO
POA
PA = PB
=
POB
Prove: PA = PB
Using Congruent Triangles
•
Some proofs require the idea of a line perpendicular to a plane.
A line and a plane are perpendicular if and only if they intersect and the line is perpendicular to
all lines in the plane that pass through the point of intersection
Example 2: Look at book for picture
given PO
plane x;
AO = BO
Statement
Reason
PO
Given
PO
plane x
OA;
PO
OB
< POA = 90; <POB = 90
< POA = < POB
AO =
BO
PO = PO
POA
PA = PB
=
POB
Prove: PA = PB
Def. of a line perpendicular to a plane
Using Congruent Triangles
•
Some proofs require the idea of a line perpendicular to a plane.
A line and a plane are perpendicular if and only if they intersect and the line is perpendicular to
all lines in the plane that pass through the point of intersection
Example 2: Look at book for picture
given PO
plane x;
AO = BO
Statement
Reason
PO
Given
PO
plane x
OA;
PO
OB
< POA = 90; <POB = 90
< POA = < POB
AO =
BO
PO = PO
POA
PA = PB
=
POB
Prove: PA = PB
Def. of a line perpendicular to a plane
Def. of
line
Using Congruent Triangles
•
Some proofs require the idea of a line perpendicular to a plane.
A line and a plane are perpendicular if and only if they intersect and the line is perpendicular to
all lines in the plane that pass through the point of intersection
Example 2: Look at book for picture
given PO
plane x;
AO = BO
Statement
Reason
PO
Given
PO
plane x
OA;
PO
OB
Def. of a line perpendicular to a plane
< POA = 90; <POB = 90
Def. of
< POA = < POB
Def. of = <‘s
AO =
BO
PO = PO
POA
PA = PB
=
POB
Prove: PA = PB
line
Using Congruent Triangles
•
Some proofs require the idea of a line perpendicular to a plane.
A line and a plane are perpendicular if and only if they intersect and the line is perpendicular to
all lines in the plane that pass through the point of intersection
Example 2: Look at book for picture
given PO
plane x;
AO = BO
Statement
Reason
PO
Given
PO
plane x
OA;
PO
OB
Def. of a line perpendicular to a plane
< POA = 90; <POB = 90
Def. of
< POA = < POB
Def. of = <‘s
AO =
Given
BO
PO = PO
POA
PA = PB
=
POB
Prove: PA = PB
line
Using Congruent Triangles
•
Some proofs require the idea of a line perpendicular to a plane.
A line and a plane are perpendicular if and only if they intersect and the line is perpendicular to
all lines in the plane that pass through the point of intersection
Example 2: Look at book for picture
given PO
plane x;
AO = BO
Statement
Reason
PO
Given
PO
plane x
OA;
PO
OB
Def. of a line perpendicular to a plane
< POA = 90; <POB = 90
Def. of
< POA = < POB
Def. of = <‘s
AO =
Given
BO
PO = PO
POA
PA = PB
=
Reflexive
POB
Prove: PA = PB
line
Using Congruent Triangles
•
Some proofs require the idea of a line perpendicular to a plane.
A line and a plane are perpendicular if and only if they intersect and the line is perpendicular to
all lines in the plane that pass through the point of intersection
Example 2: Look at book for picture
given PO
plane x;
AO = BO
Statement
Reason
PO
Given
PO
plane x
OA;
PO
OB
Def. of a line perpendicular to a plane
< POA = 90; <POB = 90
Def. of
< POA = < POB
Def. of = <‘s
AO =
Given
BO
PO = PO
POA
PA = PB
=
Reflexive
POB
Prove: PA = PB
SAS postulate
line
Using Congruent Triangles
•
Some proofs require the idea of a line perpendicular to a plane.
A line and a plane are perpendicular if and only if they intersect and the line is perpendicular to
all lines in the plane that pass through the point of intersection
Example 2: Look at book for picture
given PO
plane x;
AO = BO
Statement
Reason
PO
Given
PO
plane x
OA;
PO
OB
Def. of a line perpendicular to a plane
< POA = 90; <POB = 90
Def. of
< POA = < POB
Def. of = <‘s
AO =
Given
BO
PO = PO
POA
PA = PB
=
line
Reflexive
POB
Prove: PA = PB
SAS postulate
CPCTC
Using Congruent Triangles
A Way to Prove Two Segments or Two Angles Congruent
1)
Identify two triangles in which the two segments are
corresponding parts.
2)
Prove that the triangles are congruent
3)
State that the two parts are congruent, using the reason
Corresponding parts of congruent triangles are congruent
(CPCTC)