 thumb|right|ceva's theorem, case 1: the three lines are concurrent at a point o inside abc thumb|250 px|right|ceva's theorem, case 2: the three lines are concurrent at a point o outside abc 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, :\frac{af}{fb} \cdot \frac{bd}{dc} \cdot \frac{ce}{ea} = 1. 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 : \frac{af}{fb} \cdot \frac{bd}{dc} \cdot \frac{ce}{ea} = 1, 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 hÅ±d, 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 salkindalfred 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 : \frac{|\triangle bod|}{|\triangle cod|}=\frac{bd}{dc}=\frac{|\triangle bad|}{|\triangle cad|}. therefore, :\frac{bd}
Affine_geometry Theorems_in_plane_geometry Triangle_geometry