If your question is to show that triangle ABP is equilateral,

Let m be the mid-point of OP .i.e., M lies on the circle. (Since OP is equal to the diameter).

OM =OA

In triangle AOP,

__ |A __ =90^{0}

M is the midpoint of the hypotenuse.

=> OA = OM = AM (Mid-point of the hypotenuse is equidistant from all the vertices).

In triangle AOM,

OA = OM =AM.

=> AOM is equilateral triangle.

|__AOM__ =60^{0}

In triangle APB,

|__AOM__ =60^{0}

|__AOP__ + __|OAP__ + |__APO__ = 180^{0}

60^{0} + 90^{0} + __|APO__ = 180^{0}

|__APO__ = 180^{0} – 150^{0}

|__APO __= 30^{0}

|__APB__ = 2__|APO__

|__APB__ = 60^{0}

In APB,

AP = BP

|__APB__ = 60^{0}

=> APB is an equilateral triangle.

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