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# 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
$$PM^2 = QM × MR$$