PQR is a triangle right angled at P and M is a point on QR
Question.

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


Solution:

$\angle 1+\angle 2=\angle 2+\angle 4$ $\left(\mathrm{Each}=90^{\circ}\right)$

PQR is a triangle right angled at P

$\Rightarrow \angle 1=\angle 4$

Similarly, $\angle 2=\angle 3$. Now, this gives

$\Delta \mathrm{QPM} \sim \Delta \mathrm{PRM}$ (AA similarity)

$\Rightarrow \frac{\operatorname{ar}(\Delta Q P M)}{\operatorname{ar}(\Delta P R M)}=\frac{P M^{2}}{R M^{2}}$ (By theorem 6.7)

$\Rightarrow \frac{\frac{1}{2}(\mathrm{OM}) \times(\mathrm{PM})}{\frac{1}{2}(\mathrm{RM}) \times(\mathrm{PM})}=\frac{\mathrm{PM}^{2}}{\mathrm{RM}^{2}}\left(\begin{array}{l}\text { Area of a triangle } \\ =\frac{1}{2} \times \text { Base } \times \text { Height }\end{array}\right)$

$\Rightarrow \frac{\mathrm{OM}}{\mathrm{RM}}=\frac{\mathrm{PM}^{2}}{\mathrm{RM}^{2}}$

$\Rightarrow \mathrm{PM}^{2}=\mathrm{QM} \times \mathrm{RM}$ or $\mathrm{PM}^{2}=\mathrm{QM} \times \mathrm{MR}$
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