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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Japanese theorem for cyclic quadrilaterals ： ウィキペディア英語版 Japanese theorem for cyclic quadrilaterals In geometry, the Japanese theorem states that the centers of the incircles of certain triangles inside a cyclic quadrilateral are vertices of a rectangle. Triangulating an arbitrary cyclic quadrilateral by its diagonals yields four overlapping triangles (each diagonal creates two triangles). The centers of the incircles of those triangles form a rectangle. Specifically, let $\backslash square\; ABCD$ be an arbitrary cyclic quadrilateral and let be $M\_1,M\_2,M\_3,M\_4$ the incenters of the triangles $\backslash triangle\; ABD,\; \backslash triangle\; ABC,\; \backslash triangle\; BCD,\; \backslash triangle\; ACD$. Then the quadrilateral formed by $M\_1,M\_2,M\_3,M\_4$ is a rectangle. Note that this theorem is easily extended to prove the Japanese theorem for cyclic polygons. To prove the quadrilateral case, simply construct the parallelogram tangent to the corners of the constructed rectangle, with sides parallel to the diagonals of the quadrilateral. The construction shows that the parallelogram is a rhombus, which is equivalent to showing that the sums of the radii of the incircles tangent to each diagonal are equal. The quadrilateral case immediately proves the general case by induction on the set of triangulating partitions of a general polygon. ==See also== *Carnot's theorem *Sangaku *Wasan 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Japanese theorem for cyclic quadrilaterals」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.