Introduction:
We know that the SAS congruence criterion applies when two sides and the included angle of one triangle (respectively) equal the two sides and the included angle of another triangle.
But what if the equality is between pairs of two sides and a nonincluded angle?
Think: Do we then have something like an SSA congruence criterion?
The answer is NO! Let's try to elaborate it. Also please go through SAS congrunce criterion once to understand it in a better way.
Why SSA is not a congruence criterion?
Consider the following figure:
In the two triangles \(\Delta ABC\) and \(\Delta DEF\), we have:

\(AB = DE\)

\(BC = EF\)

\(\angle C = \angle F\) (nonincluded angles)
We see that even though two pairs of sides and a pair of angles are (correspondingly) equal, the two triangles are not congruent.
✍Note: Clearly, congruence does not hold through SSA, even though it does hold through SAS.
If you are still not convinced, let’s do an exercise. Suppose that there is a triangle two of whose sides have lengths \(4{\text{ }}cm\) and \(3{\text{ }}cm\), and a nonincluded angle is \(30^\circ \). Let’s try to geometrically construct such a triangle. If, during our construction process, we find that we can construct only one (unique) such triangle, then SSA congruence would be valid, but on the other hand, if we find that we can construct more than one such triangles, then SSA congruence would be invalid – because then two different triangles can have the same two lengths and a nonincluded angle.
Here’s a stepbystep construction:

Construct \(BC = 4{\text{ }}cm\).

Through \(C\), draw a ray \(CX\) such that \(\angle BCX = 30^\circ \).
 Now, you have to take a point \(A\) on the ray \(CX\) such that \(AB = 3{\text{ }}cm\). How many locations of \(A\) are possible? Configure your compass such that the distance between its tip and the pencil’s tip is \(3{\text{ }}cm\). Now, place the tip of the compass on \(B\), and you will find that you can mark off two points on \(CX\):
Thus, you have successfully constructed two different triangles (\(\Delta {A_1}BC\) and \(\Delta {A_2}BC\)) given a pair of sides and a nonincluded angle. What does this mean?
 A pair of sides and a nonincluded angle will not uniquely determine a triangle. In other words, congruence through SSA is invalid.
 A pair of sides and the included angle will uniquely determine a triangle. In other words, congruence through SAS is valid.