Q4 of 42 Page 220

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



⇒ Let ∠MPR = x


⇒ In Δ MPR, ∠MRP = 180-90-x


⇒ ∠MRP = 90-x


Similarly in Δ MPQ,


∠MPQ = 90-∠MPR = 90-x


⇒ ∠MQP = 180-90-(90-x)


⇒ ∠MQP = x


In Δ QMP and Δ PMR


⇒ ∠MPQ = ∠MRP


⇒ ∠PMQ = ∠RMP


⇒ ∠MQP = ∠MPR


⇒ Δ QMP ∼ Δ PMR


⇒ =


⇒ PM2 = MR × QM


Hence proved


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