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
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.
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.
GUIDED PRACTICE for Example 1 Therefore the given statement is false and ABC is not Congruent to CAD because corresponding sides are not congruent
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.
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 +
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
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 = 3 5.