If AOB is a diameter of a circle and C is a point on the circle, then AC² + BC² = AB²

TRUE

Let AB be the diameter of the circle with center O and C be any point on circle.

Since, diameter subtends a right angle to the circle,

∴ ∠ACB = 90°

Now, in right angled triangle ACB, by Pythagoras theorem, we have:

(AB)² = (AC)² + (AB)²

Thus, the statement is true.