** Two triangles can be congruent if one side and the adjacent angles of one are congruent to one side and the adjacent angles of another. ** I know the proof to this I just don't know the three isometries that are needed. Does anyone know these?

The proof I have it this:

Suppose we are given triangle ABC and triangle DEF. These two triangles have angles CAB and FDE which are congruent to each other. These triangles also have angles CBA and FED which are congruent to each other as well. Line segments AB and DE re congruent to each other. Now if either line segment AC is congruent to DF or line segment CB is congruent to EF, we would be finished by using SAS postulate. Therefore, we will assume that line segment AC is not congruent to line segment DF, and in particular that AC is greater than DF. Since AC is greater than DF, there exists a point C' which is not equal to C on line segment AC. Line segment AC' is congruent to line segment DF and , by SAS, triangle ABC' is congruent to triangle DEF. Thus angles ABC' is congruent to angle DEF. Now, by transitive property of angles congruent angle ABC' is congruent to Angles DEF. Now, by the transitive property of angles congruent angle ABC' is congruent to angle ABC, which contradicts the angles construction postulate. Therefore, line segment AC in congruent to line segment DF and triangle ABC is congruent to triangle DEF by SAS.