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

Given in $\Delta PQR,$ $P{R}^2-P{Q}^2=Q{R}^{2}andQM\perp PR$
To prove $Q{M}^2=PM\times MR$
Proof Since, $P{R}^2-P{Q}^2=Q{R}^2$
$\Rightarrow$ $P{R}^2-P{Q}^2=Q{R}^2$



So, $\Delta PQR$ is right angled triangle at Q.
In $\Delta QMRand\Delta PMQ$, $\angle M=\angle M$ $\left[Equal{90}^{\circ }\right]$
$\angle MQR=\angle QPM$ [ equal equal to ${90}^{\circ }-\angle R$]
$\therefore$ $\Delta QMR\sim \Delta PMQ$ [by AAA similarity criterion ]
Now, using property of area of similar triangles, we get
$\frac{ar\left(\Delta QMR\right)}{ar\left(\Delta PMQ\right)}=\frac{(QM{)}^2}{(PM{)}^2}$
$\Rightarrow \frac{\frac{1}{2}\times RM\times QM}{\frac{1}{2}\times PM\times QM}=\frac{(QM{)}^2}{(PM{)}^2}$
$[\because areaoftriangle=\frac{1}{2}\times base\times height]$
$\Rightarrow Q{M}^2=PM\times RM$