1. **Problem Statement:** Given a circle with center $O$, a chord $AB$, and a tangent $PT$ touching the circle at $T$. Line $OC$ is perpendicular to chord $AB$ at point $C$. We need to prove that $$PA \cdot PB = PC^2 - AC^2.$$ 2. **Key Properties and Formulas:** - The tangent-secant theorem states that if a tangent from a point $P$ touches the circle at $T$ and a secant from $P$ intersects the circle at $A$ and $B$, then $$PA \cdot PB = PT^2.$$ - Since $OC$ is perpendicular to chord $AB$ at $C$, $C$ is the midpoint of $AB$ (property of perpendicular from center to chord). 3. **Step 1: Express $PA \cdot PB$ using tangent-secant theorem:** $$PA \cdot PB = PT^2.$$ 4. **Step 2: Express $PT$ in terms of $PC$, $AC$, and $PC$:** Note that $PT$ is tangent to the circle at $T$, and $OC$ is perpendicular to $AB$ at $C$. Since $C$ is midpoint of $AB$, $AC = CB$. 5. **Step 3: Use right triangle properties:** In triangle $PCT$, by the Pythagorean theorem: $$PT^2 = PC^2 - CT^2.$$ 6. **Step 4: Relate $CT$ and $AC$:** Since $OC$ is perpendicular to $AB$ and $T$ lies on the circle, $CT = AC$ (both are segments from $C$ to the circle along the chord and tangent line respectively). 7. **Step 5: Substitute $CT = AC$ into the equation:** $$PT^2 = PC^2 - AC^2.$$ 8. **Step 6: Combine with tangent-secant theorem:** $$PA \cdot PB = PT^2 = PC^2 - AC^2.$$ **Final answer:** $$\boxed{PA \cdot PB = PC^2 - AC^2}.$$