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. Maths Questions, Similar Triangles Questions / By mathemerize Solution : Since, PM \(\perp\) QR, therefore, \(\triangle\) PQM ~ \(\triangle\) RPM \(\implies\)ย \(PM\over QM\) = \(MR\over PM\) So, \({PM}^2\) = QM.MR.