To prove a biconditional statement, the conditional statement and its converse must be proven. Start by assuming that a parallelogram $ABCD$ is a rhombus. Focus on the triangles formed when each diagonal is drawn.

A rhombus has four congruent sides. Furthermore, by the Reflexive Property of Congruence, $\overline{AC}$ is congruent to itself and $\overline{BD}$ is congruent to itself. With this information, $△ABC$ and $△ADC$ have three pairs of congruent sides. Similarly, $△ABD$ and $△CBD$ have three pairs of congruent sides. Therefore, by the Side-Side-Side Congruence Theorem, $△ABC\cong △ADC$ and $△ABD\cong △CBD.$ Since corresponding parts of congruent triangles are congruent, the corresponding angles of each pair of triangles are congruent. This means that the diagonals of the rhombus bisect pairs of opposite angles. The conditional statement has been proven.

Next, assume that the diagonals of parallelogram $ABCD$ bisect a pair of opposite angles. Consider the triangles formed when each diagonal is drawn.

Note that $\overline{AC}$ and $\overline{BD}$ are the common sides of the triangles formed. Moreover, they bisect a pair of opposite angles. Therefore, $△ABC$ and $△ADC$ have two pairs of congruent angles and one pair of congruent included sides. The same happens with $△ABD$ and $△CBD.$ From here, by the Angle-Side-Angle Congruence Theorem, $△ABC\cong △ADC$ and $△ABD\cong △CBD.$ This congruence implies that the corresponding sides are congruent. Therefore, parallelogram $ABCD$ has four congruent sides. This means that $ABCD$ is a rhombus.

The proof has been completed.