1. **Problem Statement:** Given a parallelogram PQRS with points T and M inside it such that PT = MR and PT \parallel MR, prove: (i) \(\triangle PTR \cong \triangle RMP\) (ii) \(RT \parallel PM\) and \(RT = PM\) 2. **Recall properties of parallelograms:** - Opposite sides are parallel and equal: \(PQ \parallel SR\), \(PS \parallel QR\), and \(PQ = SR\), \(PS = QR\). - Diagonals bisect each other: \(PR\) and \(QS\) intersect at midpoint. 3. **Given:** - \(PT = MR\) - \(PT \parallel MR\) 4. **To prove (i):** \(\triangle PTR \cong \triangle RMP\) - Consider triangles \(PTR\) and \(RMP\). - We know \(PT = MR\) (given). - \(PR\) is common side to both triangles. - Since \(PT \parallel MR\) and \(PR\) is a transversal, angles \(\angle PTR = \angle RMP\) (alternate interior angles). 5. **By ASA (Angle-Side-Angle) congruence criterion:** - Side \(PT = MR\) - Angle \(\angle PTR = \angle RMP\) - Side \(PR\) common Therefore, \(\triangle PTR \cong \triangle RMP\). 6. **To prove (ii):** \(RT \parallel PM\) and \(RT = PM\) - From congruence, corresponding parts of congruent triangles are equal: - \(RT = PM\) - \(\angle TRT = \angle PMP\) - Since \(PT \parallel MR\) and \(PR\) is transversal, \(\angle PTR = \angle RMP\). - Using congruence, \(RT \parallel PM\) follows because corresponding angles are equal. **Final answers:** (i) \(\triangle PTR \cong \triangle RMP\) (ii) \(RT \parallel PM\) and \(RT = PM\)