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, AC is congruent to itself and 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. c|c △ ABC - △ ADC & △ ABD - △ CBD [0.6em] AB ≅ AD & AB ≅ BC [0.5em] CB ≅ CD & AD ≅ DC [0.5em] AC ≅ AC & BD ≅ BD Therefore, by the Side-Side-Side Congruence Theorem, △ ABC ≅ △ ADC and △ ABD ≅ △ CBD. Since corresponding parts of congruent triangles are congruent, the corresponding angles of each pair of triangles are congruent. c|c △ ABC - △ ADC & △ ABD - △ CBD [0.6em] ∠ DAC ≅ ∠ BAC & ∠ ADB ≅ ∠ CDB [0.5em] ∠ DCA ≅ ∠ BCA & ∠ ABD ≅ ∠ CBD 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 AC and 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. c|c △ ABC - △ ADC & △ ABD - △ CBD [0.6em] ∠ DAC ≅ ∠ BAC & ∠ ADB ≅ ∠ CDB [0.5em] ∠ DCA ≅ ∠ BCA & ∠ ABD ≅ ∠ CBD [0.5em] AC ≅ AC & BD ≅ BD From here, by the Angle-Side-Angle Congruence Theorem, △ ABC ≅ △ ADC and △ ABD ≅ △ CBD. This congruence implies that the corresponding sides are congruent. c|c △ ABC - △ ADC & △ ABD - △ CBD [0.6em] AB ≅ AD & AB ≅ BC [0.5em] CB ≅ CD & AD ≅ DC Therefore, parallelogram ABCD has four congruent sides. This means that ABCD is a rhombus.

The proof has been completed.