In a `Delta PQR`, N is a point on PR, such that QN`bot`PR. If PN`cdot`NR=`QN^(2)`, then prove that `anglePQR=90^(@)`.

Explanation

To prove that \( \angle PQR = 90^\circ \) given that \( PN \cdot NR = QN^2 \) in triangle \( PQR \) with \( N \) being a point on \( PR \) such that \( QN \perp PR \), we can follow these steps:

### Step 1: Understand the Given Information
We have triangle \( PQR \) with point \( N \) on line segment \( PR \) such that \( QN \) is perpendicular to \( PR \). We are given that \( PN \cdot NR = QN^2 \).

### Step 2: Set Up the Angles
Let \( \angle PQR = x \) and \( \angle QRP = y \). Therefore, \( \angle QPR = 180^\circ - (x + y) \).

...