In figure A,B and C are three points on a circle with centre O such that ∠ BOC = 30° and ∠ AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.
∵ OA = OB = AB I Given
∴ ∆OAB is equilateral.
∴ ∠AOB = 60°
| The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Now, ∵ ADBC is a cyclic quadrilateral.
∴ ∠ADB + ∠ACB = 180°
| The sum of either pair of opposite angles of a cyclic quadrilateral is 180°
⇒ ∠ADB+ 30°= 180°
⇒ ∠ADB = 180° - 30°
⇒ ∠ADB = 150°.
If the non-parallel sides of a trapezium are equal, prove that it is cyclic. Prove that an isosceles trapezium is cyclic.