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

In the given figure, OP is equal to the diameter of the circle. Prove

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

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

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