In the given figure, OP is equal to the diameter of the circle. Prove that `triangleABP` is an equilateral triangle.

Explanation

Let `angleOP=angleOPB=theta`
(`because` tangents are equally inclined at the centre)
and radius of the circle be r.
Since, `angle1=90^(@)" "`(radius through point of contact is `_|_` to the tangent)
`:.` In right `triangleOAP,`
`sintheta=(OA)/(OP)=(r)/(2r)=(1)/(2)=sin30^(@)`
`:." "theta=30^(@)" "impliesangleAPB=20=2xx30^(@)=60^(@)" "`...(1)
Now, since `PA=PB" "` (length of tangents from an external point are equal)
`:." "angle2=angle3` (angles opposite to equal sides are equal)" "...(2)
In `triangleAPB,`
`angle2+angle3+angleAPB=180^(@)" "`(angle sum property)
`implies" "angle2+angle2+60^(@)=180^(@)" "`[from (1) and (2)]
`implies" "2angle2=120^(@)" "implies" "angle2=60^(@)`
`:." "angle2=angle3=60^(@)`
So, all the angles of `triangleAPB` are `60^(@).`
`:.triangleAPB` is an equilateral triangle. Hence Proved.