The radius of a circle is This circle is divided into two segments by a chord of length 2 cm. Prove that the angle subtended by this chord at any point of the major segment is 45°.

Given radius of circle OQ = OR = √2 cm

Length of chord, QR = 2 cm

In ΔOQR,

⇒ OQ² + OR² = (√2)² + (√2)²

= 2 + 2

= 4

= (QR)²

⇒ OQ² + OR² = QR²

∴ ΔOQR is a right angled triangle at ∠QOR.

We know that angle at the centre is double the angle on the remaining part of the circle.

⇒∠QOR = 2∠QPR

⇒ 90° = 2∠QPR

∴ ∠QPR = 45°

∴ The angle subtended by the chord at the point in major segment is 45°.

Hence proved