1. EXAMPLE 1
Use the SSS Congruence Postulate
Write a proof.
GIVEN
KL NL, KM NM
PROVE
KLM NLM
Proof
It is given that
KL NL and KM NM
By the Reflexive Property,
LM LN.
So, by the SSS Congruence Postulate,
KLM NLM

2. GUIDED PRACTICE
for Example 1
Decide whether the congruence statement is true. Explain your reasoning.
DFG HJK
SOLUTION
Three sides of one triangle are congruent to three sides of second triangle then the two triangle are congruent.
Side DG HK, Side DF JH,and Side FG JK.
So by the SSS Congruence postulate, DFG HJK.
Yes. The statement is true.

3. GUIDED PRACTICE
for Example 1
Decide whether the congruence statement is true. Explain your reasoning. 2.
ACB CAD
SOLUTION
BC AD
GIVEN :
PROVE :
ACB CAD
PROOF:
It is given that BC AD By Reflexive property
AC AC, But AB is not congruent CD.

4. GUIDED PRACTICE
for Example 1
Therefore the given statement is false and ABC is not
Congruent to CAD because corresponding sides
are not congruent

5. GUIDED PRACTICE
for Example 1
Decide whether the congruence statement is true. Explain your reasoning.
QPT RST 3.
SOLUTION
QT TR , PQ SR, PT TS
GIVEN :
PROVE :
QPT RST
PROOF:
It is given that QT TR, PQ SR, PT TS. So by
SSS congruence postulate, QPT RST. Yes the statement is true.

6. Standardized Test Practice
EXAMPLE 2
Standardized Test Practice
SOLUTION
By counting, PQ = 4 and QR = 3. Use the Distance Formula to find PR. d = y 2 – 1 ( ) x +

7. Standardized Test Practice
EXAMPLE 2
Standardized Test Practice = + 1 – 4 ( ) 2
– 1
(– 5
) ) PR = 4 2 +
(– 3 ) = 25 5 =
By the SSS Congruence Postulate, any triangle with side lengths 3, 4, and 5 will be congruent to
PQR.
The distance from (–1, 1) to (–1, 5) is 4. The distance from (–1, 5) to (–4, 5) is 3. The distance from (– 1, 1) to (–4, 5) is = 4 2 +
(– 3 5
(–4) – (–1) ( )
5 – 1) 25
The correct answer is A.
ANSWER

8. GUIDED PRACTICE
for Example 2 4.
has vertices J(–3, –2), K(0, –2), and L(–3, –8).
RST has vertices R(10, 0), S(10, – 3), and T(4, 0).
Graph the triangles in the same coordinate plane and show that they are congruent.
JKL
KJ = SR = 3.
ANSWER
JL = RT = 6.
LK = TS =