Question

PQR is a triangle where angle P is right angle and P and M is point on QR such that PM perpendicular to QR. then prove PM^2=QM.MR​

Answer 1

Given: ΔPQR is right angled at P is a point on QR such that PM ⊥QR.

To prove: PM2 = QM × MR

Proof: In ΔPQM, we have

PQ2 = PM2 + QM2 [By Pythagoras theorem]

Or, PM2 = PQ2 - QM2 ...(i)

In ΔPMR, we have

PR2 = PM2 + MR2 [By Pythagoras theorem]

Or, PM2 = PR2 - MR2 ...(ii)

Adding (i) and (ii), we get

2PM2 = (PQ2 + PM2) - (QM2 + MR2)

          = QR2 - QM2 - MR2        [∴ QR2 = PQ2 + PR2]

          = (QM + MR)2 - QM2 - MR2

          = 2QM × MR

∴ PM2 = QM × MR