Question

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

Answer 1

Given, $\Delta PQR$ is right angled at P and M is a point on QR such that $PM\perp QR.$
To prove that $P{M}^2=QM\times MR$
Proof :

In $\Delta PQM$,
we have $P{Q}^2=P{M}^2+Q{M}^2\left[ByPythagorastheorem\right]$
Or, $P{M}^2=P{Q}^2-Q{M}^2...\left(i\right)$

In $\Delta PMR$,
we have $P{R}^2=P{M}^2+M{R}^2$ [By Pythagoras theorem]
Or, $P{M}^2=P{R}^2-M{R}^2...\left(ii\right)$

Adding (i) and (ii), we get
$2P{M}^2=(P{Q}^2+P{R}^2)-(Q{M}^2+M{R}^2)$
$=Q{R}^2-Q{M}^2-M{R}^2[\therefore Q{R}^2=P{Q}^2+P{R}^2]$
$=(QM+MR{)}^2-Q{M}^2-M{R}^2[(a+b{)}^2=a^2+b^2+2ab$]
$=Q{M}^2+M{R}^2+2QMMR-Q{M}^2-M{R}^2$
$=2QM\times MR$
$i.e2P{M}^2=2QM\times MR$
$\therefore P{M}^2=QM\times MR$