#### Question

In a circle, with centre O, a cyclic quadrilateral ABCD is drawn with AB as a diameter of the circle and CD equal to radius of the circle. If AD and BC produced meet at point P; show that ∠APB = 60°.

#### Solution

Join OD and OC.

ΔOCD, OD = OD= CD

∴ ΔOCD is an equilateral triangle

∴ ∠ODC =60°

Also, in cyclic quadrilateral ABCD

∠ADC + ∠ABC = 180°

(pair of opposite angles in cyclic quadrilateral are supplementary)

⇒ ∠ODA + 60° + ∠ABP =180°

⇒ ∠OAD + ∠ABP = 90° ( ∵ OA = OD)

⇒ ∠PAB = ∠ABP =120°

By angle sum property of ΔPAB,

∴ ∠APB = 180° - ∠PAB -∠ABP =180° -120° =60°

Is there an error in this question or solution?

Solution In a Circle, with Centre O, a Cyclic Quadrilateral Abcd is Drawn with Ab as a Diameter of the Circle and Cd Equal to Radius of the Circle. If Ad and Bc Produced Meet at Point P; Show that ∠Apb = 60°. Concept: Chord Properties - Equal Chords Are Equidistant from the Center.