Question

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

Answer 1

Given,

$\to PQRB$ a right angle $△le$ at $P$

$\to M.B$ a point on $QR$ such that $PM\perp QR$

These we can see to right angle $△le$ namely $△PMR$ and $△PMQ$ are joined, which are right angle at $M$.

From $△lePMR$, using pythagoras theorem

$P{R}^2=P{M}^2+M{R}^2------\left(1\right)$

From $△lePMQ$, using pythagoras theorem

$P{Q}^2=P{M}^2+M{Q}^2------\left(2\right)$

Adding $\left(1\right)$ & $\left(2\right)$

$P{R}^2+P{Q}^2-2P{M}^2+M{R}^2+M{Q}^2------\left(3\right)$

But we know from $△leRPQ$, using pythagoras theorem

$P{R}^2+P{Q}^2=R{Q}^2------\left(4\right)$

Substituting $\left(4\right)$ in $\left(3\right)$

$P{Q}^2-2P{M}^2+M{R}^2+M{Q}^2$

we can see that $PQ=RM+MQ$

$\therefore (RM+MQ{)}^2=2P{M}^2+M{R}^2+M{Q}^2$

Using $(a+b{)}^2=a^2+b^2+2ab$

$\Rightarrow R{M}^2+M{Q}^2+2RM.MQ=2P{M}^2+M{R}^2+M{Q}^2$

$\Rightarrow P{M}^2=RM.MQ$

$\because$ Proved.