1. **Problem Statement:** In parallelogram ABCD, points P and Q lie on diagonal BD such that $DP = BQ$. We need to prove: (i) $\triangle APD \cong \triangle CQB$ (ii) $AP = CQ$ (iii) $\triangle AQB \cong \triangle CPD$ (iv) $AQ = CP$ (v) Quadrilateral APCQ is a parallelogram. 2. **Key Properties and Formulas:** - Opposite sides of a parallelogram are equal and parallel: $AB = DC$, $AD = BC$. - Diagonals bisect each other: midpoint of BD is also midpoint of AC. - Congruence criteria: SAS (Side-Angle-Side), ASA (Angle-Side-Angle), or SSS (Side-Side-Side). 3. **Proof of (i) $\triangle APD \cong \triangle CQB$:** - Given $DP = BQ$. - Since ABCD is a parallelogram, $AB \parallel DC$ and $AD \parallel BC$. - Angles $\angle APD$ and $\angle CQB$ are vertically opposite angles, so $\angle APD = \angle CQB$. - Also, $AD = BC$ (opposite sides of parallelogram). - By SAS criterion: $AD = BC$, $\angle APD = \angle CQB$, and $DP = BQ$ imply $\triangle APD \cong \triangle CQB$. 4. **Proof of (ii) $AP = CQ$:** - From congruence in (i), corresponding sides are equal, so $AP = CQ$. 5. **Proof of (iii) $\triangle AQB \cong \triangle CPD$:** - Similarly, $BQ = DP$ (given). - Angles $\angle AQB$ and $\angle CPD$ are vertically opposite, so equal. - $AB = DC$ (opposite sides). - By SAS, $\triangle AQB \cong \triangle CPD$. 6. **Proof of (iv) $AQ = CP$:** - From congruence in (iii), corresponding sides are equal, so $AQ = CP$. 7. **Proof of (v) APCQ is a parallelogram:** - We have $AP = CQ$ and $AQ = CP$. - Opposite sides of quadrilateral APCQ are equal. - A quadrilateral with both pairs of opposite sides equal is a parallelogram. - Therefore, APCQ is a parallelogram. **Final answers:** (i) $\triangle APD \cong \triangle CQB$ (ii) $AP = CQ$ (iii) $\triangle AQB \cong \triangle CPD$ (iv) $AQ = CP$ (v) APCQ is a parallelogram.