Question

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

Answer 1

Step 1 : Given information in the question

Given that, in $∆PQR$

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

and, $QM\perp PR$

So, in $∆PQR$

$$\begin{gathered} \Rightarrow P{R}^2-P{Q}^2=Q{R}^2 \\ \Rightarrow P{R}^2=Q{R}^2+P{Q}^2 \end{gathered}$$

Thus,$∆PQR$ is right angle triangle at $Q$.

Step 2 : Prove that $Q{M}^2=PM\times MR$

From$∆QMR$ and $∆PMQ$,

$\angle M=\angle M$

$\angle MQR=\angle QPM$ $\left[\because \angle MQR=\angle QPM=90°-\angle R\right]$

$∆QMR~∆PMQ$ [Using, $AAA$ criteria ]

Step 3: Applying the property of the triangle,

$$\begin{gathered} \Rightarrow \frac{\mathrm{Area}∆QMR}{\mathrm{Area}∆PMO}=\frac{Q{M}^2}{P{M}^2}......\left(1\right) \\ \Rightarrow \frac{\mathrm{Area}∆QMR}{\mathrm{Area}∆PMO}=\frac{{\displaystyle \frac{1}{2}}RM\times QM}{{\displaystyle \frac{1}{2}}PM\times QM}.....\left(2\right) \end{gathered}$$

From $\left(1\right)$ and $\left(2\right)$

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

Hence Proved, $Q{M}^2=PM\times MR$.