In a $\triangle\text{PQR},\text{PR}^2-\text{PQ}^2=\text{QR}^2$ and M is a point on side PR such that $\text{QM}\perp\text{PR}$ Provr that- $QM^2 = PM \times MR.$
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Given in A $PQR,$
$PR^2 - PQ^2 = QR^2$
and $\text{QM}\perp\text{PR}$
To Prove $QM^2 = PM \times MR$
Proof since, $PR^2 - PQ^2 = QR^2$
So, $\triangle\text{PQR}$ is right angled triangle at Q.
In $\triangle\text{QMR},\triangle\text{PMR}$
$\angle\text{M}=\angle\text{M}$ [each $90^\circ$]
$\angle\text{MQR}=\angle\text{QPM}$ $[\text{each equal to } 90^\circ-\angle\text{R}]$
$\therefore\triangle\text{QMR}\sim\triangle\text{PMQ}$ [by AAA similarity criterion]
Now, using property of area of similar triangles, we get
$\frac{\text{ar}(\triangle\text{QMR})}{{\text{ar}(\triangle\text{PMQ})}}=\frac{(\text{QM})^2}{(\text{PM})^2}$
$\Rightarrow\frac{\frac{1}{2}\times\text{RM}\times\text{QM}}{\frac{1}{2}\times\text{PM}\times\text{QM}}=\frac{(\text{QM})^2}{(\text{PM})^2}$ $[\because\text{area of triangle}=\frac{1}{2}\times\text{base}\times\text{height}]$
$\Rightarrow\text{QM}^2=\text{PM}\times\text{RM}$ Hence proved.
$PR^2 - PQ^2 = QR^2$
and $\text{QM}\perp\text{PR}$
To Prove $QM^2 = PM \times MR$
Proof since, $PR^2 - PQ^2 = QR^2$
So, $\triangle\text{PQR}$ is right angled triangle at Q.
In $\triangle\text{QMR},\triangle\text{PMR}$
$\angle\text{M}=\angle\text{M}$ [each $90^\circ$]
$\angle\text{MQR}=\angle\text{QPM}$ $[\text{each equal to } 90^\circ-\angle\text{R}]$
$\therefore\triangle\text{QMR}\sim\triangle\text{PMQ}$ [by AAA similarity criterion]
Now, using property of area of similar triangles, we get
$\frac{\text{ar}(\triangle\text{QMR})}{{\text{ar}(\triangle\text{PMQ})}}=\frac{(\text{QM})^2}{(\text{PM})^2}$
$\Rightarrow\frac{\frac{1}{2}\times\text{RM}\times\text{QM}}{\frac{1}{2}\times\text{PM}\times\text{QM}}=\frac{(\text{QM})^2}{(\text{PM})^2}$ $[\because\text{area of triangle}=\frac{1}{2}\times\text{base}\times\text{height}]$
$\Rightarrow\text{QM}^2=\text{PM}\times\text{RM}$ Hence proved.
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