In a circle with center O, AB is a diameter. From a point P on the circle, a perpendicular PN is drawn to AB. Prove geometrically that \[ PB^2 = AB \times BN \]

Q.In a circle with center O, AB is a diameter. From a point P on the circle, a perpendicular PN is drawn to AB. Prove geometrically that \[ PB^2 = AB \times BN \]

Given: AB is the diameter of a circle centered at O. P is any point on the circle, and PN ⊥ AB. To prove: \(PB^2 = AB \times BN\) Proof: Since AB is the diameter of the circle, ∠APB is an angle in a semicircle. ∴ ∠APB = 90°, a right angle. In the right-angled triangle APB, PN is drawn perpendicular from the vertex P to the hypotenuse AB. ∴ Triangles ∆ABP and ∆PBN are similar. So, \(\cfrac{PB}{BN} = \cfrac{AB}{PB}\) ∴ \(PB^2 = AB \times BN\) (Proved)