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Question
In ∆PQR, ∠Q = 90° and QM is perpendicular to PR. Prove that:
- PQ2 = PM × PR
- QR2 = PR × MR
- PQ2 + QR2 = PR2
Solution
i. In ∆PQM and ∆PQR,
∠PMQ = ∠PQR = 90°
∠QPM = ∠RPQ ...(Common)
∴ ∆PQM ~ ∆PRQ ...(By AA Similarity)
∴ `(PQ)/(PR) = (PM)/(PQ)`
`=>` PQ2 = PM × PR
ii. In ∆QMR and ∆PQR,
∠QMR = ∠PQR = 90°
∠QRM = ∠QRP ...(Common)
∴ ∆QRM ~ ∆PQR ...(By AA similarity)
∴ `(QR)/(PR) = (MR)/(QR)`
`=>` QR2 = PR × MR
iii. Adding the relations obtained in (i) and (ii), we get,
PQ2 + QR2 = PM × PR + PR × MR
= PR(PM + MR)
= PR × PR
= PR2
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