Question

In the given figure OP is equal to diameter of the circle.Prove thatABP is a equilateral triangle

Answer 1

The given information can be represented using a figure as

⇒ ∠OPA = 30°
Similarly, it can be proved that ∠OPB = 30°.
Now, ∠APB = ∠OPA + ∠OPB = 30° + 30° = 60°
In ∆PAB,
PA = PB [lengths of tangents drawn from an external point to a circle are equal]
⇒∠PAB = ∠PBA ...(1) [Equal sides have equal angles opposite to them]
∠PAB + ∠PBA + ∠APB = 180° [Angle sum property]
⇒∠PAB + ∠PAB = 180° – 60° = 120° [Using (1)]
⇒2∠PAB = 120°
⇒∠PAB = 60° ...(2)
From (1) and (2)
∠PAB = ∠PBA = ∠APB = 60°
∴ ∆PAB is an equilateral triangle.

Hope! This will help you.
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