Question

Two triangles QPR and QSR, right angled at P and S respectively are drawn on the same base QR and on the same side of QR. If PR and SQ intersect at T, prove that PT x TR = ST x TQ.

Answer 1

In ΔRPQ,

RP² + PQ² = QR²

∴ PQ² = QR² – RP² … (1)

In ΔTPQ,

TP² + PQ² = QT²

∴ PQ² = QT² – TP² … (2)

Equating (1) and (2) we get,

QR² – RP² = QT² – TP²

RP = RT + TP

∴ QR² – (RT + TP)² = QT² – TP²

∴ QR² – RT² – TP² – 2RT.TP = QT² – TP²

QR² = QT² + RT² + 2RT.TP … (3)

In ΔQSR,

QS² + SR² = QR²

SR² = QR² – SR² … (4)

In ΔTSR,

ST² + SR² = TR2

∴ SR² = TR² – TS² … (5)

Equating (4) and (5) we get

QR² – SQ² = TR² – TS²

SQ = QT + TS

∴ QR² – (2T + TS)² = TR² – TS²

QR² – 2T² – TS² – 2QT.TS = TR² – TS²

∴ 2R² = TR² + QT² + 2QT.TS … (6)

Now equating (3) – (6), we get

QT² + RT² + 2RT. TP = QT² + RT² + 2QT.TS

∴ PT.TR = ST.TQ

Hence proved.