Question

Given RS and PT are altitudes of triangle PQR prove that triangle Pqt ~∆QRS PQ×QS=RQ×QT

Answer 1

Answer:

Altitudes of ΔPQR = RS and PT

Altitudes of ΔPQR = RS and PT1. In ΔPQT and ΔQRS,

Altitudes of ΔPQR = RS and PT1. In ΔPQT and ΔQRS,∠PTQ = ∠RSQ = 90° (Given)

Altitudes of ΔPQR = RS and PT1. In ΔPQT and ΔQRS,∠PTQ = ∠RSQ = 90° (Given)∠PQT = ∠RQS (Common angle)

Altitudes of ΔPQR = RS and PT1. In ΔPQT and ΔQRS,∠PTQ = ∠RSQ = 90° (Given)∠PQT = ∠RQS (Common angle)ΔPQT ~ Δ RQS (By similarity AA)

Altitudes of ΔPQR = RS and PT1. In ΔPQT and ΔQRS,∠PTQ = ∠RSQ = 90° (Given)∠PQT = ∠RQS (Common angle)ΔPQT ~ Δ RQS (By similarity AA)2. Since ΔPQT and ΔRQS are similar

Altitudes of ΔPQR = RS and PT1. In ΔPQT and ΔQRS,∠PTQ = ∠RSQ = 90° (Given)∠PQT = ∠RQS (Common angle)ΔPQT ~ Δ RQS (By similarity AA)2. Since ΔPQT and ΔRQS are similarThus,

Altitudes of ΔPQR = RS and PT1. In ΔPQT and ΔQRS,∠PTQ = ∠RSQ = 90° (Given)∠PQT = ∠RQS (Common angle)ΔPQT ~ Δ RQS (By similarity AA)2. Since ΔPQT and ΔRQS are similarThus,PQ/RQ = QT/QS

Altitudes of ΔPQR = RS and PT1. In ΔPQT and ΔQRS,∠PTQ = ∠RSQ = 90° (Given)∠PQT = ∠RQS (Common angle)ΔPQT ~ Δ RQS (By similarity AA)2. Since ΔPQT and ΔRQS are similarThus,PQ/RQ = QT/QS= PQ × QS = RQ × QT

Altitudes of ΔPQR = RS and PT1. In ΔPQT and ΔQRS,∠PTQ = ∠RSQ = 90° (Given)∠PQT = ∠RQS (Common angle)ΔPQT ~ Δ RQS (By similarity AA)2. Since ΔPQT and ΔRQS are similarThus,PQ/RQ = QT/QS= PQ × QS = RQ × QTHence proved

Step-by-step explanation:

Hope it will help you ☺️. Please Mark me as brainliest ☺️.