Question

In the given figure, PQRS is a trapezium in which PQ ‖ SR and PS = QR. Prove that: ∠PSR = ∠QRS and ∠SPQ = ∠RQP

Answer 1

Construction:
Draw SM ⊥ PQ and RN ⊥ PQ

a. In ΔPMS and ΔQNR,
PS = QR                 ....(given)
∠PMS = ∠QNR    ....(Each 90°)
SM = RN               ....(Distance between parallel lines)
∴ ΔPMS ≅ ΔQNR ....(RHS congruence)
⇒ ∠PM = ∠RQN
⇒ ∠SPQ = ∠RQP
⇒ ∠PSR = ∠QRS ....(Supplement of each angle SPQ and RQP)

b. In ΔPQS and ΔQPR,  
PS = QR                ....(given)
∠SPQ = ∠RQP    ....(proved)
PQ = PQ              ....(common)
∴ ΔPQS ≅ ΔQPR ....(SAS congruence)
⇒ QS = PR          ....(proved)
⇒ PSQ = ∠QRP  ....(i)

c. In ΔTPS and ΔTQR, 
PS = QR              ....(given)
∠STP = ∠RTQ    ....(Vertically opposite angles)
∠PST = ∠QRT    ....[From (i)]
∴ ΔTPS ≅ ΔTQR ....(SAS congruence)
⇒ TP = TQ          ....(proved)
⇒ TS = TR.           ....(proved)