It is proved that in quadrilateral ABCD, connecting AC, triangle ABC and triangle CDA, using AB = CD, BC = DA, AC * * * so triangle ABC and triangle CDA are congruent, so angle BAC= angle DCA, angle DAC= angle BCA, so AB//DC and AD//BC, so quadrilateral ABCD is a parallelogram.

Is another "true proposition"? Questions like this are generally false, right? " Not necessarily.