Question

In a $△PQR,P{R}^2-P{Q}^2=Q{R}^2$ and $M$ is a point on side $PR$ such that $QM\perp PR$. Prove that $Q{M}^2=PM\times MR$.

Answer 1

Given, In $\Delta PQR$

Refer image,

$P{R}^2-P{Q}^2=Q{R}^2$

and $QM\perp PR$

To prove: $M{Q}^2=MP\times MR$

Proof: In $\Delta PQR$

$P{R}^2-P{Q}^2=Q{R}^2$ [Given]

$P{R}^2=P{Q}^2+Q{R}^2$

$\therefore \Delta PQR={90}^0$ [By conv, of Pythagoras theorem]

Now in $\Delta QMP$ and $\Delta QMR$

$\angle 1=\angle 2={90}^0$ $(\because QM\perp PR)$

$\angle P={90}^0-\angle R$

$\angle 3={90}^0-\angle R$

$\angle P=\angle 3$

$\therefore \Delta QMP\sim \Delta RMQ$ (By AA similarity criterion)

$\frac{PQ}{QR}=\frac{PM}{QM}=\frac{QM}{RM}$

$=Q{M}^2=PM\times RM$

Hence proved.