PQR is a triangle right angled at P and M is a point on QR such that PM $\perp$ QR. Show that $P{M}^2=QM.MR$.

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,
$(RQ{)}^2=(RP{)}^2+(PQ{)}^2........(3)$
$\therefore$ $(RM+MQ{)}^2=(RP{)}^2+(PQ{)}^2$
$\therefore$ $(RM{)}^2+(MQ{)}^2+2RM.MQ=(RP{)}^2+(PQ{)}^2....(4)$

Adding 1) and 2) we get,
$(PR{)}^2+(PQ{)}^2=2(PM{)}^2+(RM{)}^2+(MQ{)}^2...(5)$

From 4) and 5) we get,
$2RM.MQ=2(PM{)}^2$
$\therefore$ $(PM{)}^2=RM.MQ$