Question

in the given figure, op is equal to the diameter of a circle with centre o and pa and pb are tengents .prove that abp is an equilateral triangle

Answer 1

In the figure, Where OP intersects with circle, Mark it as 'Q'
now,

OQ = radius 
OP = 2 radius  (diameter)
QP is also radius
and,
angle OAP and angle ABP are 90 degrees (due to point where tangent meet)

We know that,
angle in a semi-circle is 90 degrees,

hence,
A circle passes through O, A, P, B

AQ = QB = radius
AO = OB = radius

and,
AQ is also equal to AO
AQ = QB = BO = OA = radius

hence,
quad.AQBO is a rhombus.

We know that,
opposite angles of rhombus are equal

angle.AOB = angle.AQB

In circle with centre 'Q'
angle. AQB = 2(angle. APB)
[angle subtended by arc at the centre is double the angle subtended by it at any point on remaining part of the circle]

in quad. APBO,
=> angle. AOB + 90 + 90 + angle. APB = 360 degrees
=> 2(angle. APB) + 90 + 90 + angle APB. = 360 degress
=> 3(angle. APB) = 180 degrees
=> angle. APB = 60 degrees

In triangle. APB,
AP = PB (tangents from an external point are equal)

hence, 
triangle. APB is isoceles
then,
angle. PAB = angle. PBA

now,
=> angle. PAB + angle. PBA + angle. APB = 180 degrees
=> 2(angle. PAB) + 60 = 180 degrees
=> 2(angle. PAB) = 120 degrees
=> angle. PAB = 60 degrees

angle. PAB = angle. PBA = angle. APB = 60 degrees

hence,
triangle APB is equilateral triangle