Must it be a parallelogram?

If we draw a diagonal $BD$ and show that $BD = BD$ by the reflexive property, then we have two triangles $\triangle ABD$ and $\triangle CDB$ that have two pairs of congruent sides and one pair of congruent angles.

Unfortunately, SSA is not a theorem to show two triangles congruent, so we cannot conclude that $AD = CB$ to show that $ABCD$ must be a parallelogram. In fact, by the law of sines on both triangles, $\sin \angle DBC = \frac{AB \sin \angle A}{BD} = \sin \angle BDA$ , but since $\sin x = \sin (180° - x)$ , either $\angle DBC = \angle BDA$ or $\angle DBC = 180° - \angle BDA$ .

Now, if we add the condition that $\angle DBC = 90°$ , then by either of the above parameters $\angle BDA = 90°$ , so the two triangles are congruent by AAS congruency, which means $AD = CD$ , and so $ABCD$ is a parallelogram. Therefore, the choice "if $\angle DBC = 90°$ , then $ABCD$ is a parallelogram" is always true.

For completeness, we can find counter-examples to the other choices. The following quadrilateral has $\angle DAB = \angle BCD = 60°$ , $\angle ABD = 50°$ , $\angle ADB = 70°$ , $\angle DBC = 110°$ , and $\angle BDC = 10°$ ; and meets the given conditions that $AB = CD$ and $\angle DAB = \angle DCB$ , but $\angle DBC \neq \angle BDA$ , so opposite angles $\angle ABC \neq \angle ADC$ which makes it not a parallelogram. Therefore, $ABCD$ is not necessarily a parallelogram, so the choice " $ABCD$ is a parallelogram" is not always true.

Finally, the following parallelogram is made up of two equilateral triangles, and also meets the given conditions that $AB = CD$ and $\angle DAB = \angle DCB$ , but $\angle DBC = 60° \neq 90°$ . Therefore, "if $ABCD$ is a parallelogram, then $\angle DBC = 90°$ " is not always true.