If an equilateral triangle be inscribed in a circle, the square on the side of the triangle is triple of the square on the radius of the circle.
若在圆内接一个等边三角形，则该三角形一边上的正方形面积等于圆半径上的正方形面积的三倍。
正三角形 ABC 内接于圆 O；D 是弧 BC 的中点（A 对侧），E 是 BC 中点。证明 AB² = 3·OA² 即三角形边长平方等于半径平方的三倍。
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Let ABC be a circle, and let the equilateral triangle ABC be inscribed in it; I say that the square on one side of the triangle ABC is triple of the square on the radius of the circle. For let the centre D of the circle ABC be taken, let AD be joined and carried through to E, and let BE be joined.
设圆ABC，内接等边三角形ABC，圆心D。连接AD并延长至E，连接BE。
Then, since the triangle ABC is equilateral, therefore the circumference BEC is a third part of the circumference of the circle ABC. Therefore the circumference BE is a sixth part of the circumference of the circle; therefore the straight line BE belongs to a hexagon; therefore it is equal to the radius DE.
因三角形ABC等边，弧BEC为圆周的三分之一，故弧BE为六分之一，线段BE为正六边形的一边，等于半径DE。
[IV. 15, Por.] And, since AE is double of DE, the square on AE is quadruple of the square on ED, that is, of the square on BE. But the square on AE is equal to the squares on AB, BE; [III. 31, I. 47] therefore the squares on AB, BE are quadruple of the square on BE.
因AE是DE的二倍，故AE上的正方形是ED上的正方形的四倍，即BE上的正方形的四倍。
Therefore, separando, the square on AB is triple of the square on BE. But BE is equal to DE; therefore the square on AB is triple of the square on DE.
但AE上的正方形等于AB、BE上的正方形之和，故AB、BE上的正方形之和是BE上的正方形的四倍，分离后得AB上的正方形是BE上的正方形的三倍。因BE等于DE，故AB上的正方形是DE上的正方形的三倍。