In a triangle $ P Q R, N $ is a point on $ P R $ such that $ Q N \perp P R $. If $ P N $. $ N R=Q^{2} $, prove that $ \angle \mathrm{PQR}=90^{\circ} $.

Given:

In a triangle \( P Q R, N \) is a point on \( P R \) such that \( Q N \perp P R \).

\( P N \). \( N R=Q^{2} \)

To do:

We have to prove that \( \angle \mathrm{PQR}=90^{\circ} \).

Solution:

$P N . N R=Q N^{2}$

$P N . N R=Q N . Q N$

This implies,

$\frac{P N}{Q N}=\frac{Q N}{N R}$...........(i)

In $\triangle QNP$ and $\triangle RNQ$,

$\frac{P N}{Q N}=\frac{Q N}{N R}$

$\angle P N Q=\angle R N Q$

Therefore, by SAS similarity,

$\angle Q N P \sim \angle R N Q$

This implies,

$\angle P Q N=\angle Q R N$........(ii)

$\angle R Q N=\angle Q P N$........(iii)

Adding (ii) and (iii), we get,

$\angle P Q N+\angle R Q N=\angle Q R N+\angle Q P N$

$\angle P Q R=\angle Q R N+\angle Q P N$

We know that,

Sum of the angles of a triangle is \( 180^{\circ} \).

In $\triangle P Q R$, 

$\angle P Q R+\angle Q P R+\angle Q R P=180^{\circ}$

$\angle P Q R+\angle Q P N+\angle Q R N=180^{\circ}$

$\angle P Q R+\angle P Q R=180^{\circ}$

$2 \angle P Q R=180^{\circ}$

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

Hence proved.