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