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定理 设三角形的Ｂｒｏｃａｒｄ角是θ，外接圆半径是Ｒ，则正负Ｂｒｏｃａｒｄ点间的距离是２Ｒ１－４ｓｉｎ２θ·ｓｉｎθ．引理１ 将△ＡＢＣ绕外心Ｏ反时针旋转２θ得△Ａ１Ｂ１Ｃ１，则△ＡＢＣ的正Ｂｒｏｃａｒｄ点与△Ａ１Ｂ１Ｃ１的负Ｂｒｏｃａｒｄ点重合．图１证
Theorem Set the Brocade angle of the triangle to θ and the radius of the circumcircle to R. Then the distance between positive and negative Brocard points is 2R1-4sin2θ·sinθ. Lemma 1 rotates ΔABC by 2θ counterclockwise around the center O to obtain ΔA1B1C1, then the positive Brocard point of ΔABC coincides with the negative Brocard point of ΔA1B1C1. Figure 1 card