To prove that SSA condition does not prove congruence, we need one counter example. We will try to make a triangle made of three points \(A\), \(B\) and \(C\) such that \(A\) and \(B\) being 5 units apart, \(A\) and \(C\) being 3 units apart and \(\angle ABC\) being \(30^\circ\).
Consider points \(A\) and \(B\) that are 5 units apart and that lie on the \(x\)-axis:
Let point \(C\) be 3 units away from \(A\). The location of \(C\) is unknown, so it lies somewhere on this circle:
Let there be a red line touching point \(B\) and raised \(30^\circ\) above the \(x\)-axis:
To make a triangle out of \(A\), \(B\) and \(C\), such that \(A\) and \(B\) are 5 units apart, \(A\) and \(C\) are 3 units apart and \(\angle ABC\) is \(30^\circ\), we would need point \(C\) to lie on one of the two intersections of the circle:
This means with our given conditions, we can make two triangles:
This example shows that a side-side-angle condition doesn't necessarily give us a unique triangle.