In \(\triangle PQR\), point \(N\) lies on \(PR\) with \(QN\perp PR\). If \(PN\cdot NR = QN^2\), prove that \(\angle PQR = 90^\circ\).
Proved.
Step 1: We are given a triangle \(\triangle PQR\) where \(N\) is a point on side \(PR\), and \(QN\) is drawn perpendicular to \(PR\).
Step 2: The condition is \(PN \cdot NR = QN^2\).
Step 3: Recall an important property (called the geometric mean theorem): In a right triangle, if you drop a perpendicular from the right angle to the hypotenuse, then: \[(\text{Altitude})^2 = (\text{Segment 1 of hypotenuse}) \times (\text{Segment 2 of hypotenuse}).\]
Step 4: Here, \(QN\) is the altitude, \(PN\) and \(NR\) are the two segments of \(PR\). The condition \(QN^2 = PN \cdot NR\) exactly matches this property.
Step 5: Since this property holds only when \(PR\) is the hypotenuse of a right triangle, it means \(\triangle PQR\) must be a right-angled triangle.
Step 6: Therefore, the right angle must be at vertex \(Q\). So, \(\angle PQR = 90^\circ\).
Final Answer: Proved.