**Theorem 6.8**: (SAS) Two
triangles are
congruent if and only if two
sides and the
angle between them in one
triangle are
congruent to the two
sides and the
angle between them in the
other triangle.

**Proof**: If the two
triangles are
congruent, then when the whole
triangle is moved to
the other triangle, the
three angles and three
sides of the one
triangle will be moved
to the corresponding parts of the other
triangle and they will
all have to be congruent. As a
result, any two sides
and the angle between them
in one triangle will be
congruent to the corresponding
two sides and the
angle between them in a
second triangle.

For the converse, le* * *A*,* **B*,* *and* **C *be the vertices of one
triangle and* **A*',* **B*',* *and* **C*'* *be the vertices
of the other triangle
such that

and

Translate * **A*'* *to* **A*. * *Rotate the
plane about
point * **A *until* **B*'* *lies on the ray from* **A *to* **B*.* *At this
point, since both* **B *and* **B*'* *will be on the same line, the same
distance from* **A *in the
same direction, by Theorems
2.1 and 2.2, * **B*' will
coincide with* **B*. Since* **m*__/ __*A* = *m*__/ __*A*',* *by Theorem 5.13, there are two
possibilities for the line
segment * **A*'*C*':* *it is either coincident with the line determined by* **AC *or it is
on the reflection of that
line about* **AB*, by Theorem 4.12. If it is the reflection,
reflect back about* **AB *and by Theorem 4.9, it will
coincide with* **AC*.* *So after a translation,
rotation, and, perhaps,
a reflection,* **C*'* *will be on
the line determined by* **AC*.* *Since* **C*'* *will be the same distance from* **A *as* **C*,* *we
must conclude that* **C*'* *is moved to* **C*.* *Thus* *triangle *A*'*B*'*C*'* *is moved to* *triangle * ABC *by an invertible isometry, and the
triangles are
congruent.