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.

Solution :

Since, PM \(\perp\) QR, therefore,

\(\triangle\) PQM ~ \(\triangle\) RPM

\(\implies\)  \(PM\over QM\) = \(MR\over PM\)

So, \({PM}^2\) = QM.MR.