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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 Parallelogram law ： ウィキペディア英語版 Parallelogram law In mathematics, the simplest form of the parallelogram law (also called the parallelogram identity) belongs to elementary geometry. It states that the sum of the squares of the lengths of the four sides of a parallelogram equals the sum of the squares of the lengths of the two diagonals. Using the notation in the diagram on the right, the sides are (''AB''), (''BC''), (''CD''), (''DA''). But since in Euclidean geometry a parallelogram necessarily has opposite sides equal, or (''AB'') = (''CD'') and (''BC'') = (''DA''), the law can be stated as, : $2(AB)^2+2(BC)^2=(AC)^2+(BD)^2\backslash ,$ In case the parallelogram is a rectangle, the two diagonals are of equal lengths (''AC'') = (''BD'') so, : $2(AB)^2+2(BC)^2=2(AC)^2\backslash ,$ and the statement reduces to the Pythagorean theorem. For the general quadrilateral with four sides not necessarily equal, : $(AB)^2+(BC)^2+(CD)^2+(DA)^2=(AC)^2+(BD)^2+4x^2.\backslash ,$ where ''x'' is the length of the line joining the midpoints of the diagonals. It can be seen from the diagram that, for a parallelogram, ''x'' = 0, and the general formula is equivalent to the parallelogram law. ==The parallelogram law in inner product spaces== In a normed space, the statement of the parallelogram law is an equation relating norms: :$2\backslash |x\backslash |^2+2\backslash |y\backslash |^2=\backslash |x+y\backslash |^2+\backslash |x-y\backslash |^2.\; \backslash ,$ In an inner product space, the norm is determined using the inner product: :$\backslash |x\backslash |^2=\backslash langle\; x,\; x\backslash rangle.\backslash ,$ As a consequence of this definition, in an inner product space the parallelogram law is an algebraic identity, readily established using the properties of the inner product: :$\backslash |x+y\backslash |^2=\backslash langle\; x+y,\; x+y\backslash rangle=\; \backslash langle\; x,\; x\backslash rangle\; +\; \backslash langle\; x,\; y\backslash rangle\; +\backslash langle\; y,\; x\backslash rangle\; +\backslash langle\; y,\; y\backslash rangle,\; \backslash ,$ :$\backslash |x-y\backslash |^2\; =\backslash langle\; x-y,\; x-y\backslash rangle=\; \backslash langle\; x,\; x\backslash rangle\; -\; \backslash langle\; x,\; y\backslash rangle\; -\backslash langle\; y,\; x\backslash rangle\; +\backslash langle\; y,\; y\backslash rangle.\; \backslash ,$ Adding these two expressions: :$\backslash |x+y\backslash |^2+\backslash |x-y\backslash |^2\; =\; 2\backslash langle\; x,\; x\backslash rangle\; +\; 2\backslash langle\; y,\; y\backslash rangle\; =\; 2\backslash |x\backslash |^2+2\backslash |y\backslash |^2,\; \backslash ,$ as required. If ''x'' is orthogonal to ''y'', then $\backslash langle\; x\; ,\backslash \; y\backslash rangle\; =\; 0$ and the above equation for the norm of a sum becomes: :$\backslash |x+y\backslash |^2=\; \backslash langle\; x,\; x\backslash rangle\; +\; \backslash langle\; x,\; y\backslash rangle\; +\backslash langle\; y,\; x\backslash rangle\; +\backslash langle\; y,\; y\backslash rangle\; =\backslash |x\backslash |^2+\backslash |y\backslash |^2,$ which is Pythagoras' theorem. 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「Parallelogram law」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.