In a circle of radius 5 cm,AB and AC are two chords such that AB = AC = 6 cm.Find the length of chord BC.
Given AB and AC are two equal words of a circle, therefore the centre of the circle lies on the bisector of ∠BAC.
⇒ OA is the bisector of ∠BAC.
Again, the internal bisector of an angle divides the opposite sides in the ratio of the sides containing the angle.
∴ P divides BC in the ratio = 6 : 6 = 1 : 1.
⇒ P is mid-point of BC.
⇒ OP ⊥ BC.
In ΔABP, by pythagoras theorem,
In right triangle OBP, we have
Equating (1) and (2), we get
Putting AP in (1), we get