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Find the area of the triangle inside the circle

Question: An equilateral triangle inscribed in a circle. The radius of the circle is r and center of the circle is point O. Find the area of the equilateral triangle.

Extended AO, which intersects BC at point D, and intersects the circle at point E.
Because ABC is an equilateral triangle, so each interior angle is 60o, so angle C = 60o
Because the point C is on the circle, so the angle C is an inscribed angle.
Because angle C is opposite the arc AB and the angle E is also opposite the arc AB,
so, angle E = angle C (Inscribed angles opposite the same arc are equal.)
Connect BO, because BO = EO = r,
so, angle OBE = angle E = 60o (In a triangle, congruent sides opposite congruent angles.)
so, angle BOE = 180o - 60o - 60o = 60o
so, triangle BOE is an equilateral triangle (In a triangle, if each interior angle is 60o, then the triangle is an equilateral triangle.)
so, BE = r
Because AE is a diameter, so angle ABE = 90o (In a circle, the diameter opposite 90o angle.)
In the right triangle ABE,
AB2 + BE2 = AE2
b2 + r2 = (2r)2
so, b2 = 4r2 - r2 = 3r2
so, b = (square root of 3) r
The area (A) of the triangle ABC is,
A = (1/2)BC × AD
= (1/2) b × (square root of 3) b/2
= (1/4) × b2 × (square root of 3)
= (1/4) × 3r2 × (square root of 3)
= 3(square root of 3)r2/4
so, the area of the inscribed triangle is equal to 3(square root of 3)r2/4.