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Ceva's theorem is a theorem about triangles in Euclidean plane geometry. Given a triangle ''ABC'', let the lines ''AO'', ''BO'' and ''CO'' be drawn from the vertices to a common point ''O'' to meet opposite sides at ''D'', ''E'' and ''F'' respectively. (The segments ''AD, BE,'' and ''CF'' are known as cevians.) Then, using signed lengths of segments, : In other words the length ''AB'' is taken to be positive or negative according to whether ''A'' is to the left or right of ''B'' in some fixed orientation of the line. For example, ''AF''/''FB'' is defined as having positive value when ''F'' is between ''A'' and ''B'' and negative otherwise. The converse is also true: If points ''D'', ''E'' and ''F'' are chosen on ''BC'', ''AC'' and ''AB'' respectively so that : then ''AD'', ''BE'' and ''CF'' are concurrent. The converse is often included as part of the theorem. The theorem is often attributed to Giovanni Ceva, who published it in his 1678 work ''De lineis rectis''. But it was proven much earlier by Yusuf Al-Mu'taman ibn Hűd, an eleventh-century king of Zaragoza. Associated with the figures are several terms derived from Ceva's name: cevian (the lines AD, BE, CF are the cevians of O), cevian triangle (the triangle DEF is the cevian triangle of O); cevian nest, anticevian triangle, Ceva conjugate. (''Ceva'' is pronounced Chay'va; ''cevian'' is pronounced chev'ian.) The theorem is very similar to Menelaus' theorem in that their equations differ only in sign. ==Proof of the theorem== (Here directed segments are not used, except in the case of proving using Menelaus' Theorem) A standard proof is as follows;〔Follows Russel〕 Posamentier and Salkind〔Alfred S. Posamentier and Charles T. Salkind, ''Challenging Problems in Geometry'', Dover Publishing Co., second revised edition, 1996.〕 give four proofs. First, the sign of the left-hand side is positive since either all three of the ratios are positive, the case where ''O'' is inside the triangle (upper diagram), or one is positive and the other two are negative, the case ''O'' is outside the triangle (lower diagram shows one case). To check the magnitude, note that the area of a triangle of a given height is proportional to its base. So : Therefore, : (Replace the minus with a plus if ''A'' and ''O'' are on opposite sides of ''BC''.) Similarly, : and : Multiplying these three equations gives : as required. The theorem can also be proven easily using Menelaus' theorem.〔Follows 〕 From the transversal ''BOE'' of triangle ''ACF'', : and from the transversal ''AOD'' of triangle ''BCF'', : The theorem follows by dividing these two equations. The converse follows as a corollary.〔Follows Russel〕 Let ''D'', ''E'' and ''F'' be given on the lines ''BC'', ''AC'' and ''AB'' so that the equation holds. Let ''AD'' and ''BE'' meet at ''O'' and let ''F''′ be the point where ''CO'' crosses ''AB''. Then by the theorem, the equation also holds for ''D'', ''E'' and ''F''′. Comparing the two, : But at most one point can cut a segment in a given ratio so ''F''=''F''′. 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「Ceva's theorem」の詳細全文を読む スポンサード リンク
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