Transcript 4.4 Proving Congruence – SSS, SAS

4.2 Triangle Congruence
by SSS, SAS


SOL: G6
Objectives: The Student Will …
Use the SSS Postulate to test for triangle
congruence
Use the SAS Postulate to test for triangle
congruence
Side-Side-Side Congruence
If the three sides of one triangle are
congruent to the three sides of another
triangle, then the two triangles are
congruent.
 Abbreviation: SSS
If AB  DE,

BC  EF,
AC  DF,
Then ABC  DEF
Examples: Are the two triangles congruent
1.)
A F
G
L
2.)
M
C
B
Given:
AB  EF
AC  EG
BC  FG
Then by
SSS Congruence
ABC  EFG
E
N
Since
LM  ON
LN  OM
O
MN  NM –
Reflexive Property
Then by
SSS Congruence
LMN  ONM
Included Angle

The angle formed by two sides
D
B
F
E
A
C
B is the Included Angle
of sides AB and CB
D is the Included Angle
of sides FD and ED
Side-Angle-Side Congruence

If two sides and the included angle of one
triangle are congruent to two sides and the
included angle of another triangle, then the
two triangles are congruent.
A is the included angle
of sides AB and AC.
D is the included angle
of sides DE and DF.
If AB  DE, AC  DF, ∡A  ∡D,
then ABC  DEF
Example 3:
Determine which postulate can be used to prove
that the triangles are congruent. If it is not
possible to prove that they are congruent, write
not possible.
SSS
SAS
Example 4:
What other information do you
need to prove ∆LEB  ∆BNL by SAS?
Given:
∡ELB  ∡NBL
Reflexive Property:
LB  BL
We still need the 2nd side that makes
the included angle:
LE  BN
Example 5:
Would you use SSS or SAS to prove the triangles
below congruent?
SSS
we can not use SAS since it is not the included angle,
look at the markings on the sides that make the
angle
Example 6: Are the two triangles congruent
Given:
BC and DE bisect each other
BG  CG, and DG  EG
DGB  EGC, vertical angles
SAS Congruence, then DGB  EGC
Example 7:
Determine whether WDV  MLP.
We can find the length of the sides by using the
Pythagorean theorem or the distance formula.
WD  ( 7  5)  ( 4  1)  4  9  13

2

2
DV  ( 5  1) ( 1  2)

2

 16  1  17
2
VW  ( 1  7)  ( 2  4)  36  4  40

2

2
ML  (4  1)  ( 7  5)

2
LP  (1  2)  ( 5  1)

2
W(-7, -4)
D(-5, -1)
V(-1, -2)
M(4, -7)
L(1, -5)
P(2, -1)
 9  4  13
2
 1 16  17
2
PM  (2  4) ( 1  7)  4  36  40
2

2
Since the corresponding sides are
Congruent then WDV  MLP, by SSS
Homework

Pg 230 – 231 # 9 – 15, 18 – 20, 21