In ΔPQR,∠PQR=90° So, a circle can be drawn passing through the points P, Q and R with PR as diameter. ∴ PT = QT = TR(Radius of the circle) In ΔQTR QT = TR ⇒ ∠TRQ = ∠TQR..........(1) LM || QT ∴∠RLM = ∠TQR......(2) (Corresponding angles) From (1) and (2), we have ∠RLM= ∠LRM (∠TRQ = ∠LRM)