PQR is a triangle right angled at P and M is a point on QR such that PM \(\perp\) QR. Show that \(PM^2 = QM × MR\).

Given, \(\triangle\) PQR is right angled at P is a point on QR such that PM ?QR



We have to prove, \(PM^2 = QM × MR\)

In \(\triangle\) PQM, by Pythagoras theorem
\(PQ^2 = PM^2 + QM^2\)
Or, \(PM^2 = PQ^2 – QM^2\) ……………………………..(i)

In \(\triangle\) PMR, by Pythagoras theorem
\(PR^2 = PM^2 + MR^2\)
Or, \(PM^2 = PR^2 – MR^2\) ………………………………………..(ii)

Adding equation, (i) and (ii), we get,
\(2PM^2 = (PQ^2 + PM^2) – (QM^2 + MR^2)\)
= \(QR^2 – QM^2 – MR^2\) [Thus \(QR^2 = PQ^2 + PR^2\)]
= \((QM + MR)^2 – QM^2 – MR^2\)
= 2QM × MR
\(PM^2 = QM × MR\)