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Transcript
Geometry 4-4 Proving Triangles Congruent - SSS, SAS We saw last time that, in order to show two triangles are congruent, we need to show that all three sets of corresponding sides and all three sets of corresponding angles are congruent. There are ways to show two triangles congruent without showing all six corresponding parts congruent. Postulate 4.1 - Side-Side-Side (SSS) Congruence: If the three sides of one triangle are congruent to the three sides of a second triangle, then the two triangles are congruent. The angle formed by two adjacent sides of a polygon is called an included angle. Postulate 4.2 - Side-Angle-Side (SAS) Congruence: If two sides and the included angle of one triangle are congruent to two sides and the included angles of a second triangle, then the triangles are congruent. Triangle JKL has vertices J(2,5), K(1,1), and L(5,2). Triangle NPQ has vertices N(-3,0), P(-7,1), and Q(-4,4). Are the triangles congruent? Since there are JK = (2 - 1)2 + (5 - 1)2 = 1 + 16 = 17 three sets of KL = (1 - 5)2 + (1 - 2)2 = 16 + 1 = 17 congruent sides, 2 2 LJ = (2 - 5) + (5 - 2) = 9 + 9 = 18 the triangles are NP = (-3 - -7)2 + (0 - 1)2 = 16 + 1 = 17 congruent. PQ = (-7 - -4)2 + (1 - 4)2 = 9 + 9 = 18 QN = (-4 - -3)2 + (4 - 0)2 = 1 + 16 = 17 JKL ~ = NPQ Write a paragraph proof. Given: BC ~ = DC, BCF~ = DCE, FC ~ = EC ~ Prove: CFD= CEB B F E D ~ = We are given BC ~ = DC, BCF ~ = DCE, FC ~ = EC. By the SAS Postulate , BCF ~ = DCE. Since the C triangles are congruent , we can say BFC~ = DEC. Now, BFC and EFC form a linear pair, as do DEC and FEC. So CFD ~ = CEB by the Congruent Supplements Theorem. CPCTC: Corresponding Parts of Congruent Triangles are Congruent