12. If \(S\) is a point on side \(PQ\) of \(\triangle PQR\) such that \(PS=QS=RS\), then
\(PR\cdot QR = RS^2\)
\(QS^2 + RS^2 = QR^2\)
\(PR^2 + QR^2 = PQ^2\)
\(PS^2 + RS^2 = PR^2\)
Step 1: It is given that \(PS = QS = RS\). This means the point \(S\) is the same distance from all three vertices \(P, Q, R\).
Step 2: A point that is equidistant from all three vertices of a triangle is called the circumcentre of the triangle.
Step 3: Here, \(S\) lies on the side \(PQ\). Normally, the circumcentre of a triangle lies inside (for an acute triangle), outside (for an obtuse triangle), or on the midpoint of the hypotenuse (for a right triangle).
Step 4: Since \(S\) lies on side \(PQ\), the only possibility is that the triangle \(PQR\) is a right-angled triangle with the right angle at \(R\).
Step 5: In a right-angled triangle, the side opposite the right angle is called the hypotenuse. Here, side \(PQ\) is the hypotenuse.
Step 6: By the Pythagoras theorem, we know that:
\[ PR^2 + QR^2 = PQ^2 \]
Final Answer: Option (C) is correct.