Fig. 6.343. In Fig. 6.35. RT andOR QTOS PR. In Fig. 6.35, andRnd 1 42.

1.	Fig. 6.343. In Fig. 6.35. RT andOR QTOS PR. In Fig. 6.35, andRnd 1 42. Show thatFig. 6.35
Answer» In ΔPQR, ∠1 = ∠2∠PQR = ∠PRQ [GIVEN]∴ PR = PQ ……………..…(1) [Sides opposite to equal angles of a triangle are also equal]Given: QR/QS = QT/PRQR/QS = QT/PQ QS/QR = PQ/QT…... …(ii) [Taking reciprocals] [From eq (i)]In ΔPQS and ΔTQR,QS/QR = PQ/QT [From eq (ii)] ∠PQS = ∠TQR [ common]∴ ΔPQS ~ ΔTQR [By SAS similarity criterion] Hence, proved Like my answer if you find it useful!

Fig. 6.343. In Fig. 6.35. RT andOR QTOS PR. In Fig. 6.35, andRnd 1 42. Show thatFig. 6.35

Answer»

In ΔPQR, ∠1 = ∠2∠PQR = ∠PRQ [GIVEN]∴ PR = PQ ……………..…(1)

[Sides opposite to equal angles of a triangle are also equal]Given: QR/QS = QT/PRQR/QS = QT/PQ

QS/QR = PQ/QT…... …(ii) [Taking reciprocals]

[From eq (i)]In ΔPQS and ΔTQR,QS/QR = PQ/QT [From eq (ii)] ∠PQS = ∠TQR [ common]∴ ΔPQS ~ ΔTQR [By SAS similarity criterion]

Hence, proved

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