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As shown in the figure, in △ ABC, the bisector of the outer angle ∠ CBD and ∠ BCE intersects at point O, and it is proved that ∠ BOC = 90 ° - & frac12; ∠ a
According to the meaning of the title: ﹥ a + ﹥ ABC + ﹥ ACB = 180 °, ﹥ OBC = 1 / 2 ﹥ CBD = 1 / 2 (﹥ a + ﹥ ACB), ﹥ OCB = 1 / 2 ﹥ BCE = 1 / 2 (﹥ a + ﹥ ABC), and ﹥ BOC = 180 ° - (﹥ OBC + ﹥ OCB), so ﹥ BOC = 180 ° - [1 / 2 ﹥ a + 1 / 2 (﹥ a + ﹥ ACB + ﹥ ABC)] = 180 ° - (1 / 2 ﹥ a + 90 ')
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