From a point P, two tangents PA and PB are drawn to a circle with center O. If OP =
diameter of the circle shows that ΔAPB is equilateral.
OP = 2r
Tangents drawn from external point to the circle are equal in length
PA = PB
At point of contact, tangent is perpendicular to radius.
In ΔAOP, sin 𝜃 =`"opp.side"/"hypotenuse"=r/(2r)=1/2`
𝜃 = 30°
∠APB = 20 = 60°, as PA = PB ∠BAP = ∠ABP = x.
In ΔPAB, by angle sum property
∠APB + ∠BAP + ∠ABP = 180°
2x = 120° ⇒ x = 60°
In this triangle all angles are equal to 60°
∴ ΔAPB is equilateral.