Start with an isosceles triangle, say, an equilateral triangle, with all sides
having measure, say, of 10:



Now draw a slanted line from A to D where D is NOT the midpoint
of BC, so that AD has measure, say, 7 and DB has measure 3:




S:   AB and AC have the same measure, 10
S:   The green line AD has the same measure as itself
A:   Angle B and angle C have the same measure of 60°.

However triangles ABD and ACD are NOT congruent.

Comments:

1. You can do this with any isosceles triangle, by drawing a line
   from the vertex of the vertex angle to a point on the base other
   than its midpoint.

2. This version of the SSA theorem can be proved:

SSA Theorem: If two sides and a nonincluded angle of one triangle are
congruent to the corresponding two sides and angle of a second triangle, then
the triangles are either congruent or else the angles opposite the longer
congruent sides are supplementary.  

3.  With regard to 2, we can see in the above case that the two angles 
at D are supplementary.


Edwin