Answer:
SSA is not a valid congruence theorem
Step-by-step explanation:
Given a figure with opposite sides the same length, and their ends joined by a line through the midpoint of the segment joining their other ends, you want to know why the shape is not a parallelogram.
SSA congruence
The recognized congruence theorems are ...
AAS, ASA, SAS, SSS
The "would be" SSA theorem is not included, because the case where the angle is opposite the shortest of the two sides does not uniquely specify the triangle.
In this diagram, the sides marked with 2 and 1 hash marks are sequential with the vertical angle where the diagonals cross. Hence the only basis for claiming the triangles congruent is SSA.
The basis for claiming congruence is not a recognized congruence theorem, so we cannot conclude the triangles are congruent. (CPCTC does not apply.)
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Additional comment
As we noted, the SSA criterion for congruence fails if the given angle cannot be demonstrated to be opposite the longest side. The attachment shows the case where the single-hash segment is longer than the double-hash segment, so there are two ways to draw the figure. One is a parallelogram, the other is not.
The HL congruence theorem is essentially SSA, where the A is the right angle. If the angle can be demonstrated to be the largest in the triangle, then the SSA theorem can be used (as for right- or obtuse triangles).
As the attachment shows, there is nothing in this geometry that requires the vertical angle(s) opposite the double hash be the largest.
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