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In this chapter, we will discuss another congruence theorem, this time the Side-Angle-Side theorem.
The angle is called the included angle.

== The Side-Angle-Side congruence theorem ==
Given two triangles <math>\triangle ABC </math> and <math>\triangle DEF </math> such that their sides are equal, hence:
# The side <math>\overline {AB} </math> equals <math>\overline {DE} </math>.
# The side <math>\overline {CA} </math> equals <math>\overline {DF} </math>.
# The angle <math>\angle CAB </math> equals <math>\angle FDE </math> (These are the angles between the sides).

Then the triangles are congruent and their other angles and sides are equal too.  Success!

== Proof ==
We will use the method of superposition – we will move one triangle to the other one and we will show that they coincide.
We won’t use the construction we learnt to copy a line or a segment but we will move the triangle as whole.

# <math>\text{Superpose } \triangle ABC \text{ on } \triangle DEF \text{ such that } A \text { is placed on } D \text { and } \overline {AB} \text{ is placed on } \overline {DE}</math>
# <math>\because \overline {AB} = \overline {DE} \text{ (given)}</math>
# <math>\therefore B \text{ coincides with } E </math>
# <math>\because \angle CAB = \angle FDE \text{ (given)}</math>
# <math>\therefore \overline {CA} = \overline {FD} </math>
# <math>\because \overline {CA} = \overline {DF} \text{ (given)}</math>
# <math>\therefore C \text{ coincides with } F </math>
# <math>\therefore \overline {CB} = \overline {EF} </math>
# <math>\therefore \triangle ABC = \triangle DEF </math>
# <math>\therefore \triangle ABC \cong \triangle DEF </math>

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