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In a triangle PQR, N is a point on PR such that QN $\perp$ PR . If PN. NR = QN², prove that $\angle$ PQR = 90° .

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Given:- In a triangle PQR, N is a point on PR such that QN $\perp$ PR and PN.NR = QN²

To prove:- $\angle$ PQR = 90°

Proof:

We have PN.NR = QN²

$\\\Rightarrow PN.NR=QN.QN\\\frac{PN}{QN}=\frac{QN}{NR}\; \; \; \; \; \; \; ....(1)$

$In \Delta QNP \; and\; \Delta RNQ$

$\frac{PN}{QN}=\frac{QN}{NR}$

and $\angle$ PNQ = $\angle$ RNQ (each equal to 90°)

we know that if one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio, then the triangles are similar by SAS similarity criterion.

$\therefore \Delta QNP \sim \Delta RNQ$

Then $\Delta$ QNP and $\Delta$ RNQ are equiangular.

i.e. $\angle PQN = \angle QRN \; \; \; \; \; ....(2)$

$\angle RQN = \angle QPN \; \; \; \; \; ....(3)$

adding equation (2) and (3) we get

$\angle PQN + \angle RQN = \angle QRN + \angle QPN$

$\Rightarrow \angle PQR = \angle QRN + \angle QPN \; \; \; \; \; ...(4)$

In triangle PQR

$\angle PQR + \angle QPR + \angle QRP = 180^{o}$

$\angle PQR + \angle QPN + \angle QRN = 180^{o}$

$\begin{Bmatrix} \because \angle QPR=\angle QPN \\\angle QRP=\angle QRN \end{Bmatrix}$

$\angle PQR + \angle PQR = 180^{o}$

[Using equation (4)]

$2\angle PQR=180^{o}$

$\angle PQR=\frac{180^{o}}{2}$

$\angle PQR=90^{o}$

Hence proved

Posted by

infoexpert23

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