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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:
PQR is a triangle right angled at P and M is a point on QR such that $PM \perp QR$.
To do:
We have to show that $PM^2 = QM • MR$.
Solution:
In right angle triangle $PQR$,
$\angle P = 90^o$
$PM \perp QR$
In $\triangle \mathrm{PMR}$, by Pythagoras theorem
$(\mathrm{PR})^{2}=(\mathrm{PM})^{2}+(\mathrm{RM})^{2}$..........(i)
In $\triangle \mathrm{PMQ}$, by Pythagoras theorem,
$(PQ)^{2}=(P M)^{2}+(M Q)^{2}$..........(ii)
In $\triangle \mathrm{PQR}$, by Pythagoras theorem,
$(R Q)^{2}=(R P)^{2}+(P Q)^{2}$............(iii)
Therefore,
$(RM+MQ)^{2}=(R P)^{2}+(P Q)^{2}$
$(R M)^{2}+(M Q)^{2}+2 R M . M Q=(R P)^{2}+(P Q)^{2}$.........(iv)
Adding (i) and (ii), we get
$(\mathrm{PR})^{2}+(\mathrm{PQ})^{2}=2(\mathrm{PM})^{2}+(\mathrm{RM})^{2}+(\mathrm{MQ})^{2}$.........(v)
From (iv) and (v), we get,
$2 \mathrm{RM} . \mathrm{MQ}=2(\mathrm{PM})^{2}$
$(\mathrm{PM})^{2}=\mathrm{RM} . \mathrm{MQ}$
Hence proved.