In fig Op is equal to diameter of the circle. Prove that abp is a equilateral triangle..
sry abt the figure but i dnt know how to upload the figure
But its a circle with two tangents AP and BP... and joining OP, OA and OB ( o is the centre of the circle... )
AP is the tangent to the circle.
∴ OA ⊥ AP (Radius is perpendicular to the tangent at the point of contact)
⇒ ∠ OAP = 90º
In Δ OAP,