6. Theorem of Ceva, Menelaus and Van Aubel.

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and so we conclude that the point A ′′ on the line BC coincides with the point A1. Thus the points A1, B1 and C1 are collinear. Definition 1 A line segment joining a vertex of a triangle to any given point on the opposite side is called a Cevian. Theorem 2 (Ceva) Three Cevians AA1, BB1 and CC1 of a triangle ABC (Figure 2) are concurrent if and only if |BA1| |A1C| .|CB1| |AC1| . |B1A| |C1B| = 1. Proof First assume that the Cevians are concurrent at the point M. Consider the triangle AA1C and apply Menelaus’ theorem. Since the points B1, M and B are collinear, |B1C| |MA| . |B1A| |MA1| .|BA1| |BC| = 1 . . . (a) Now consider the triangle AA1B. The points C1, M, C are collinear so |C1A| |CB| . |C1B| |CA1| .|MA1| = 1 . . . (b) |MA| Multiply both sides of equations (a) and (b) to get required result. Figure 2: Conversely, suppose the two Cevians AA1 and BB1 meet at P and that the Cevian from the vertex C through P meets side AB at C ′ . Then we have By hypothesis, Thus |BA1| |A1C| .|CB1| |B1A| . |AC′ | |C ′ B| = 1. |BA1| |A1C| .|CB1| |AC1| . = 1. |B1A| |C1B| |AC1| |C1B| = |AC′ | |C ′ B| , and so the two points C1 and C ′ on the line segment AB must coincide. The required result follows. 2
Theorem 3 (van Aubel) If A1, B1, C1 are interior points of the sides BC, CA and AB of a triangle ABC and the corresponding Cevians AA1, BB1 and CC1 are concurrent at a point M (Figure 3), then |MA| |MA1| |C1A| |B1A| = + |C1B| |B1C| . Proof Again, as in the proof of Ceva’s theorem, we apply Menelaus’ theorem to the triangles AA1C and AA1B. In the case of AA1C, we have and so |B1A| |B1C| |B1C| |MA| . |B1A| |MA1| .|BA1| |BC| |MA| = |MA1| .|BA1| |BC| For the triangle AA1B, we have and so |C1A| |C1B| Adding (c) and (d) we get as required. Examples |B1A| |C1A| + |B1C| |C1B| = |C1A| |CB| . . |C1B| |CA1 |MA1| |MA| |MA| = |MA1| .|CA1| |BC| = 1, = 1, . . . (c) . . . (d) |MA| |MA1||BC| {|BA1| + |A1C|} = |MA| |MA1| , 1. Medians AA1, BB1 and CC1 intersect at the centroid G and then since 1 = |A1B| |A1C| |GA| |GA1| = 2, = |B1C| |B1A| 3 = |C1A| |C1B| . Figure 3:
Page 1: 6. Theorem of Ceva, Menelaus and Va
Page 5 and 6: Figure 5: Multiplying the 3 ratios,
Page 7 and 8: Theorem 4 Let ABC be a triangle wit
Page 9 and 10: Then Proof Let x = |AX| and y = |AY

6. Theorem of Ceva, Menelaus and Van Aubel.

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