Moderate 649 likes

Question

AB is the diameter of the circle with centre O. From a point P on the circle, a perpendicular PN is drawn on AB. Prove geometrically that PB² = AB.BN.

WBBSE Class X Mathematics Theorem Related To Angle In A Circle 05 Mark 2025

Answer

Given: AB is the diameter of a circle with center O. From a point P on the circle, a perpendicular PN is drawn on AB.

AB is the diameter of the circle with centre O. From a point P on the circle a perpendicular PN is drawn on ABTo Prove: PB2 = AB × BN

Proof: AB is the diameter, so ∠ APB = 90º

PN is perpendicular to AB.

Triangles PNB and APB are similar by AA similarity.

By similarity property,

\frac{PB}{AB} = \frac{BN}{PB}

Cross multiplying,

PB2 = AB × BN  (Hence proved)

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