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Question
Given: RS and PT are altitudes of ΔPQR. Prove that:
- ΔPQT ~ ΔQRS,
- PQ × QS = RQ × QT.
Solution
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
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