Question

In ∆PQR, ∠Q = 90° and QM is perpendicular to PR. Prove that:

1. PQ² = PM × PR
2. QR² = PR × MR
3. PQ² + QR² = PR²

Answer 1

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