Given: AC is the diameter of a circle centered at O, ABC is an inscribed triangle, and OP ⊥ AB (where P lies on the circle). **Prove that:** \[ OP : BC = 1 : 2 \]

Q.Given: AC is the diameter of a circle centered at O, ABC is an inscribed triangle, and OP ⊥ AB (where P lies on the circle). **Prove that:** \[ OP : BC = 1 : 2 \]

Given: O is the center of a circle, and AC is its diameter. Triangle ABC is inscribed in the circle, and OP ⊥ AB. To Prove: \(OP : BC = 1 : 2\) Proof: In triangle ∆ABC, \(\angle ABC = 90^\circ\) [Angle in a semicircle is a right angle] Also, \(\angle OPA = 90^\circ\) [Since OP ⊥ AB] ∴ OP is parallel to BC Again, O is the midpoint of AC [Since O is the center and AC is the diameter] Now, according to geometry, if a line is drawn from the midpoint of one side of a triangle parallel to another side, it bisects the third side and is half its length. ∴ \[ OP = \frac{1}{2} BC \Rightarrow \frac{OP}{BC} = \frac{1}{2} \] Hence, \[ OP : BC = 1 : 2 \quad \text{[Proved]} \]