QQuestion
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.
✓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.
To 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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