Transcript Proving Triangles are Congruent: SSS and SAS

Proving Triangles are
Congruent:
SSS and SAS
Chapter 4.3
Goal 1: SSS & SAS Congruence Postulates
Postulate 19: (SSS) Side-Side-Side
Congruence Postulate
 If three sides of one triangle are congruent to
three sides of a second triangle, then the two
If Side
PQ  WX,
triangles
are congruent.
Angle S  Z, and
Side Q S  XY,
If Side AB  DE,
IfSide
Side
AB
DE,
DE,
Side
BC
ED,
and
If
AB

then Side
PQ
S  EF,
WXY.
BC
and
Side
BC
ED,
and
SideBC
CAED,
FD,and
Side
Side CA
CA FD,
FD,
Side
then ABC  DEF
then ABC
ABC DEF
DEF
then
B
C
F
D
A
E
Proof
Given : Diagram
Pr ove : PQW  TSW
T
P
Q
W
Statement
1) PQ  TS
2) PW  TW
3) QW  SW
4)PQW  TSW
S
Re ason
1) Given
2) Given
3) Given
4) Side  Side  Side(SSS)
Postulate 20: (SAS) Side-Angle-Side
Congruence Postulate
 If two sides and the included angle of one triangle are
congruent to two sides and the included of a second
triangle,
then the two triangles are congruent.
If Side
PQ  WX,
Angle S  Z, and
Side Q S  XY,
If Side PQ  WX,
then
PQ
 YZ,
WXY.
P
If Side
QS
Angle
S
 SZ,
and
IfIf Side
Side PQ
PQ WX,
WX,
Side
Q S  XY,
Angle
Angle S
S  Z,
Z,and
and
Side
Side Q
QSS XY,
XY,
then Side
PQ
S

WXY.
PS XZ,
then
then
PQS
PQ
SS
PQ
XYZ
WXY.
WXY.
Q
S
X
Y
Z
Proof
E
G iven: C is the midpoint of AE and BD.
Prove: ABC  EDC
B
C
D
A
Statement
C is the midpoint of AE and BD.
Reason
G iven
AC  EC
Def. of Midpoint
BC  DC
Def. of Midpoint
ACB  ECD
ABC 
EDC
Vertical Angles
SA S
Goal 2: Modeling a Real Life Situation
Example 3: Choosing Which Congruence
Postulate to Use
Q
G iven: RQ  RS and Q P  SP
Prove: RQ P RSP
S
P
Paragraph Proof
R
The marks on the diagram show that PQ  PS and QR 
SR. By the Reflexive Property of Congruence, RP  RP.
Because the sides of ΔPQR are congruent to the
corresponding sides of ΔPSR, you can use the SSS
Congruence Postulate to prove that the triangle are
congruent.
Example 6: Congruent Triangles
in a Coordinate Plane
 Use the SSS Congruence Postulate to show
that
ABC

FGH.
AC = FH = 3
AB = FG = 5
AB
FG

A(-7,5)
H(6,5)
C(-4,5)
**Use the Distance
Formula to find the lengths
BC and GH**
Who remembers the
distance formula?
F(6,2)
G(1,2)
B(-7,0)
x2  x1    y2  y1 
2
BC = GH = √34
All sides congruent
2