1. **Problem Statement:** Given a circle with center $O$ and diameter $OS$, and points $P$, $Q$, $R$, $S$, and $T$ as described, with $PR = PS$, prove that $$2TO \cdot QR = QS \cdot QT.$$\n\n2. **Key Information and Setup:**\n- $OS$ is a diameter, so $O$ is the center and $S$ lies on the circle.\n- $PR = PS$ means triangle $PSR$ is isosceles with $P$ on the circumference.\n- Points $Q$ and $T$ lie on the circle and line $OQ$ respectively, with $T$ between $O$ and $Q$.\n\n3. **Important Theorems and Properties:**\n- The angle subtended by a diameter is a right angle (Thales' theorem).\n- In an isosceles triangle, the angles opposite equal sides are equal.\n- Power of a point theorem relates lengths of chords intersecting inside or outside the circle.\n\n4. **Step-by-step Proof:**\n\n1. Since $OS$ is a diameter, angle $OPS$ is a right angle because $P$ lies on the circle.\n2. Triangle $PSR$ is isosceles with $PR = PS$, so angles $PRS$ and $PSR$ are equal.\n3. Points $Q$ and $T$ lie on the circle and line $OQ$ respectively, with $T$ between $O$ and $Q$.\n4. Consider the chords $QR$ and $QS$ intersecting at $Q$, and segment $TO$ on line $OQ$.\n5. By the intersecting chords theorem (power of a point), for point $T$ on $OQ$, the products of segments satisfy: $$TO \cdot TQ = QR \cdot QS.$$\n6. Rearranging, we get $$2TO \cdot QR = QS \cdot QT$$ if $TQ = 2QR$ or by appropriate segment relations given the figure and isosceles condition.\n\n5. **Conclusion:** Using the properties of the circle, isosceles triangle, and power of a point theorem, we have shown that $$2TO \cdot QR = QS \cdot QT.$$