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

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Given:

In a $\Delta \mathrm{PQR}, \mathrm{PR}^{2}-\mathrm{PQ}^{2}=\mathrm{QR}^{2}$ and $\mathrm{M}$ is a point on side $\mathrm{PR}$ such that $\mathrm{QM} \perp \mathrm{PR}$.

To do:

We have to prove that $\mathrm{QM}^{2}=\mathrm{PM} \times \mathrm{MR}$.

Solution:

$\mathrm{PR}^{2}-\mathrm{PQ}^{2}=\mathrm{QR}^{2}$

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

This implies,

$\triangle P Q R$ is a right angled triangle with right angle at $Q$.

In $\triangle Q M R$ and $\triangle P M Q$,

$\angle M =\angle M=90^{\circ}$

$\angle M Q R =\angle Q P M=90^{\circ}-\angle R$

Therefore, by AA similarity,

$\triangle Q M R \sim \triangle P M Q$

Using the property of area of similar triangles, we get, $\frac{\operatorname{ar}(\Delta Q M R)}{\operatorname{ar}(\Delta P M Q)} =\frac{(Q M)^{2}}{(P M)^{2}}$

$\frac{\frac{1}{2} \times R M \times Q M}{\frac{1}{2} \times P M \times Q M}=\frac{(Q M)^{2}}{(P M)^{2}}$

$Q M^{2} =P M \times R M$

Hence proved.

Updated on 10-Oct-2022 13:28:00