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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 \anglePNQ = \angleRNQ          (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 \DeltaQNP and \DeltaRNQ 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

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