Question

Given: RS and PT are altitudes of ΔPQR. Prove that:

1. ΔPQT ~ ΔQRS,
2. PQ × QS = RQ × QT.

Answer 1

i.
In ∆PQT and ∆QRS,

∠PTQ = ∠RSQ = 90° ...(Given)

∠PQT = ∠RQS  ...(Common)

∆PQT ~ ∆RQS     ...(By AA similarity)

ii.

Since, triangle PQT and RQS are similar

∴ `(PQ)/(RQ) = (QT)/(QS)`

`=>` PQ × QS = RQ × QT