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Menelaus' theorem, named for Menelaus of Alexandria, is a theorem about triangles in plane geometry. Given a triangle ''ABC'', and a transversal line that crosses ''BC'', ''AC'' and ''AB'' at points ''D'', ''E'' and ''F'' respectively, with ''D'', ''E'', and ''F'' distinct from ''A'', ''B'' and ''C'', then : or simply : This equation uses 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 ''D'', ''E'' and ''F'' are collinear. The converse is often included as part of the theorem. The theorem is very similar to Ceva's theorem in that their equations differ only in sign. == Proof == A standard proof is as follows:〔Follows Russel〕 First, the sign of the left-hand side will be negative since either all three of the ratios are negative, the case where the line DEF misses the triangle (lower diagram), or one is negative and the other two are positive, the case where DEF crosses two sides of the triangle. (See Pasch's axiom.) To check the magnitude, construct perpendiculars from ''A'', ''B'', and ''C'' to the line ''DEF'' and let their lengths be ''a, b,'' and ''c'' respectively. Then by similar triangles it follows that |''AF''/''FB''| = |''a''/''b''|, |''BD''/''DC''| = |''b''/''c''|, and |''CE''/''EA''| = ''c''/''a''. So : For a simpler, if less symmetrical way to check the magnitude,〔Follows 〕 draw ''CK'' parallel to ''AB'' where ''DEF'' meets ''CK'' at ''K''. Then by similar triangles : and the result follows by eliminating ''CK'' from these equations. The converse follows as a corollary.〔Follows Russel with some simplification〕 Let ''D'', ''E'' and ''F'' be given on the lines ''BC'', ''AC'' and ''AB'' so that the equation holds. Let ''F''′ be the point where ''DE'' 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「Menelaus' theorem」の詳細全文を読む スポンサード リンク
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