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ContentElementary Geometry
Elementary GeometryMiscellaneous PropositionsCollinear and Concurrent Systems of Points and Lines967DefinitionsPoints lying in the same straight line are collinear. Straight lines passing through the same point are concurrent, and the point is called the focus of the encil of lines.TheoremIf the sides of the triangle 𝐴𝐵𝐶, or the sides produced, be cut by any straight line in the points 𝑎, 𝑏, 𝑐 respectively, the line is called a transversal, and the segments of the sides are connected by the equation. 968 (𝐴𝑏∶𝑏𝐶)(𝐶𝑎∶𝑎𝐵)(𝐵𝑐∶𝑐𝐴)=1 Conversely, if this relation holds, the points 𝑎, 𝑏, 𝑐 will be collinear.ProofThrough any vertex 𝐴 draw 𝐴𝐷 parallel to the opposite side 𝐵𝐶, to meet the transversal in 𝐷, then 𝐴𝑏∶𝑏𝐶=𝐴𝐷∶𝐶𝑎 and 𝐵𝑐∶𝑐𝐴=𝑎𝐵∶𝐴𝐷VI. 4 which proves the theorem.Note: In the formula the segments of the sides are estimated positive, independently of direction, the sequence of the letters being preserved the better to assist the memory. A point may be supposed to travel from 𝐴 over the segments 𝐴𝑏, 𝑏𝐶, ⋯ continuously, until it reaches 𝐴 again. 969 By the aide of (701) the above relation may be put in the form ( ProofBy the transversal 𝐵𝑏 to the triangle 𝐴𝑎𝐶. we have (968) (𝐴𝑏∶𝑏𝐶)(𝐶𝐵∶𝐵𝑎)×(𝑎𝑂∶𝑂𝐴)=1 And, by the transversal 𝐶𝑐 to the triangle 𝐴𝑎𝐵, (𝐵𝑐∶𝑐𝐴)(𝐴𝑂∶𝑂𝑎)×(𝑎𝐶∶𝐶𝐵)=1 Multiply these equations together. 971 If 𝑏𝑐, 𝑐𝑎, 𝑎𝑏, in the last figure, be produced to meet the sides of 𝐴𝐵𝐶 in 𝑃, 𝑄, 𝑅, then each of the nine lines in the figure will be divided harmonically, and the points 𝑃, 𝑄, 𝑅, will be collinear.Proof
CorIf in the same figure 𝑞𝑟, 𝑟𝑝, 𝑝𝑞 be joined, the three lines will pass through 𝑃, 𝑄, 𝑅 respectively.ProofTake 𝑂 as a focus to the triangle 𝑎𝑏𝑐, and employ (970) and the harmonic division of 𝑏𝑐 to show that the transversal 𝑟𝑞 cuts 𝑏𝑐 in 𝑃. 972 If a transversal intersects the sides 𝐴𝐵, 𝐵𝐶, 𝐶𝐷, ⋯ of any polygon in the points 𝑎, 𝑏, 𝑐, ⋯ in order, then (𝐴𝑎∶𝑎𝐵)(𝐵𝑏∶𝑏𝐶)(𝐶𝑐∶𝑐𝐷)(𝐷𝑑∶𝑑𝐸)⋯=1ProofDivide the polygon into triangles by lines drawn from one of the angles, and, applying (968) to each triangle, combine the results. 973 Let any transversal cut the sides of a triangle and their three intersectors 𝐴𝑂, 𝐵𝑂, 𝐶𝑂 (see figure of 970) in the points 𝐴′, 𝐵′, 𝐶′, 𝑎′, 𝑏′, 𝑐′, respectively; then, as before, (𝐴′𝑎′∶𝑎′𝐶′)(𝐶′𝑎′∶𝑎′𝐵′)(𝐵′𝑐′∶𝑐′𝐴′)=1ProofEach side forms a triangle with its intersector and the transversal. Take the four remaining lines in succession for transversal to each triangle, applying (968) symmetrically, and combine the twelve equations. 974 If the lines joining corresponding vertices of two triangles 𝐴𝐵𝐶, 𝑎𝑏𝑐 are concurrent, the points of intersection of the pairs of corresponding sides are collinear, and conversely.ProofLet the concurrent lines 𝐴𝑎, 𝐵𝑏, 𝐶𝑐 meet in 𝑂. Take 𝑏𝑐, 𝑐𝑎, 𝑎𝑏 transversals respectively to the triangles 𝑂𝐵𝐶, 𝑂𝐶𝐴, 𝑂𝐴𝐵, applying (968), and the product of the three equations shows that 𝑃, 𝑅, 𝑄 lie on a transversal to 𝐴𝐵𝐶. 975 Hence it folllows that, if the lines joining each pair of corresponding vertices of any two rectilineal figures are concurrent, the pairs of corresponding sides intersect in points which are collinear.The figures in this case are said to be in perspective, or in homology, with each other. The point of concurrence and the line of collineaity are called respectively the centre and axis of perspective or homology. See (1083). Sources and Referenceshttps://archive.org/details/synopsisofelementaryresultsinpureandappliedmathematicspdfdrive©sideway ID: 210900025 Last Updated: 9/25/2021 Revision: 0 Ref: References
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