Question

AB and AP are two tangents to the same circle having centre O. If OP is equal to the diamtere of the circle. Prove that ABP is an equilateral traingle.

Answer 1

AP and BP are two tangents os a circle with centre O
Let radius of circle = r
OA = OB = r
Join O to P and A to B
OP = diameter of circle so
OP = 2r

In Triangle OAP
Angle OAP = 90
(Tangent is perpendicular to the radius of circle through the point of contact )
Sin angle APO = perpendicular / hypotenuse
Sin angle APO = OA /PO
Sin angle APO = r /2r
Sin angle APO = 1/2
Sin angle APO = sin 30
Angle APO = 30

Similarly we can prove angle BPO = 30

Angle APB = angle APO + angle BPO
30 +30=60

AP = BP
(Tangents dream from the same external point are equal )
Angle ABP = angle BAP
(Angles opposite of equal sides are equal)

In triangle ABP
Angle ABP + angle BAP+ angle APB = 180

Angle ABP + angle ABP+ 60 = 180
2Angle ABP = 120
Angle ABP = 60
As
Angle ABP = angle BAP

60 = angle BAP
Hence all angles of this triangle is 60 so it is a equilateral triangle