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

|AB| = |A'B'|,

|AC| = |A'C'|,

and

m/ A = m/ A'.

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.

next theorem (6.9)