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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