PQR is a triangle right angled at P and M is a point on QR such that PM⊥QR. Show that PM²=QM.MR

In △PMR, By Pythagoras theorem, (PR)²=(PM)²+(RM)².. (1)
In △PMQ, By Pythagoras theorem, (PQ)²=(PM)²+(MQ)².. (2)
In △PQR, By Pythagoras theorem, (RQ)²=(RP)²+(PQ)² (3)
∴(RM+MQ)²=(RP)²+(PQ)²
∴(RM)²+(MQ)²+2RM.MQ=(RP)²+(PQ)² (4)
Adding 1) and 2) we get, (PR)²+(PQ)²=2(PM)²+(RM)²+(MQ)².. (5)
From 4) and 5) we get, 2RM.MQ=2(PM)²
∴(PM)²=RM.MQ