# 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.