No, it is not.
See SSA
> "Side-Side-Angle condition: If two sides and a corresponding non-included angle of a triangle have the same length and measure, respectively, as those in another triangle, then this is not sufficient to prove congruence; but if the angle given is opposite to the longer side of the two sides, then the triangles are congruent. The Side-Side-Angle condition does not by itself guarantee that the triangles are congruent because one triangle could be obtuse-angled and the other acute-angled."
Regarding your drawing, you may consider a circle centered in $C$ with radius $CB$ : it will intersect the side $AB$ into $B_1$ with $CB=CB_1$.