Given any inscribed triangle T in the circle that has at least one vertex V not "diametrically" opposite its opposite side (which is a chord of the circle), the area of T can be increased by moving the vertex V to this point on the circle that is "diametrically" opposite this chord. This is because given a side S of the triangle, the area is proportional to the perpendicular distance of the opposite vertex V from this side (using A = (1/2)bh), and from the geometry of a circle this perpendicular distance h is clearly maximal when V is "diametrically" opposite this side S.
Thus for the maximal-area inscribed triangle T, it must be one where each vertex is "diametrically" opposite its opposite side. In other words, all the altitudes are also the sides' medians, which implies this is an equilateral triangle. (*)
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By a vertex V being "diametrically" opposite its opposite side S, I mean the line connecting the midpoint M of S and the vertex V passes through the circle's centre. By HSC circle geometry theorems, this is equivalent to saying MV is the perpendicular bisector of S (use the theorem "line through centre to midpoint of chord is its perpendicular bisector" and the converse of this). So MV is both a median and an altitude.
(*) To show a triangle where the medians are also the altitudes is equilateral, you only really need to show the following result: "Let T be a triangle where one of its medians is also the altitude. Then T is isosceles." Then apply this result twice to show (*).
To show this isosceles result, you can do it by proving certain sub-triangles are congruent, using SAS (draw a diagram and it should be clear).
