If twice the square of the diameter of a circle is equal to half the sum of the squares of the sides of inscribed $\Delta$ABC, then $sin^2 A + sin^2 B + sin^2 C$ is equal to?

Using the extended sine rule, the sides of the triangle are a = 2R sin A, b = 2R sin B, and c = 2R sin C, where R is the circumradius. Since the diameter d is equal to 2R, the given equation translates to 2 times (2R)^2 = 0.5 times ((2R sin A)^2 + (2R sin B)^2 + (2R sin C)^2). Simplifying this yields 8R^2 = 2R^2 times (sin^2 A + sin^2 B + sin^2 C), which results in sin^2 A + sin^2 B + sin^2 C = 4.

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