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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.

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