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

(i) PQ² = PM × PR

(ii) QR² = PR × MR

(iii) PQ² + QR² = PR²

△PQR is shown in the figure below:

(i) In △PQR and △PMQ,

⇒ ∠PMQ = ∠PQR [Both = 90°]

⇒ ∠QPM = ∠RPQ [Common]

∴ △PQR ~ △PMQ [By AA]

Since, corresponding sides of similar triangles are proportional we have :

⇒ $\dfrac{PQ}{PR} = \dfrac{PM}{PQ}$

⇒ PQ² = PM × PR

Hence, proved that PQ² = PM × PR.

(ii) In △QRM and △PRQ,

⇒ ∠QMR = ∠PQR [Both = 90°]

⇒ ∠QRM = ∠QRP [Common]

∴ △QRM ~ △PRQ [By AA]

Since, corresponding sides of similar triangles are proportional we have :

⇒ $\dfrac{QR}{PR} = \dfrac{MR}{QR}$

⇒ QR² = PR × MR

Hence, proved that QR² = PR × MR.

(iii) Adding equations from (i) and (ii) we get,

⇒ PQ² + QR² = PM × PR + PR × MR ………(1)

⇒ PQ² + QR² = PR(PM + MR)

From figure,

PM + MR = PR

⇒ PQ² + QR² = PR².

Hence, proved that PQ² + QR² = PR².