(Q13) PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM.MR.

Solution :

Given : In ∆PQR, ∠P = 90° and PM ⊥ QR.

R.T.P :² = .

Proof :In ∆PQR; ∆MPR

∠P = ∠ [each 90°]

∠R = ∠ [common]

∴ ∆PQR ∼ ∆MPR .....(1)

[A.A. similarity]

In ∆PQR and ∆MQP,

∠P = ∠ [each 90°]

∠Q = ∠ [common]

∴ ∆PQR ∼ ∆MQP .....(2)

[A.A. similarity]

From (1) and (2),

∆PQR ∼ ∆MPR ∼ ∆MQP [transitive property]

∴ ∆MPR ∼ ∆MQP

[Ratio of corresponding sides of similar triangles are equal]

PM . = MR .

² = QM . MR