CEVA’S THEOREM elegantly describes why a point of perspectivity, joined to the vertices of a triangle, divides the opposite sides in such proportions that the product of corresponding left hand segments matches the product of corresponding right hand segments and, conversely, that if we have such a division of sides by Cevian lines through the vertices the lines must meet in a (Cevian) point.
[WHEREAS we will generally have concerns about the exact description of ‘lengths’ of segments in planar situations, provided all points remain rational (number pairs) we will have rational proportions. Regarded as displacements only therefore, our segments may still be compared as ratios – which is all we rely on when applying the Theorem.]
WE NOW take advantage of this fact to establish the remarkable result – sometimes referred to as a ‘Cevian Nest‘ – and here termed the ‘Cevian of Cevian Rule‘ that when a second Cevian triangle is ‘nested’ inside a first that its vertices (in contact with the sides of the first Cevian triangle) may be joined to the corresponding vertices of the original given triangle to give three lines which meet in a new Cevian point!
A TRIANGLE ABC (above) contains a Cevian point X with associated Cevian triangle A’B’C’. In turn, A’B’C’ contains a Cevian point Y with Cevian triangle (sides not shown) A”B”C”. If sides AB and CA are divided in the affine proportions β : (1-β) and α : (1-α) at C’ and B’ respectively then, since
it follows from Ceva’s Theorem that side BC is divided in proportion
CA’ : A’B = (αβ)/(1-α-β+2αβ) : (1-α)(1-β)/(1-α-β+2αβ) at A’.
LIKEWISE if sides B’C’, C’A’ of triangle A’B’C’ are divided in affine proportions
B’A” : A”C’ = γ : (1-γ) and
C’B” : B”A’ = δ : (1-δ) it follows that
A’C” : C”B’ = (1-γ)(1-δ)/(1 – γ -δ + 2γδ) : (cδ)/(1 – γ -δ + 2γδ)
NOW the line AA” being a barycentric combination of A and A” necessarily contains B’ and C’ in fixed proportion. Whilst every point on AA” will contain varying proportions of both A and C and A and B respectively, only the proportion of C (in B’) and B (in C’) determine how AA” divides BC (where the weight of A becomes ‘0’.) Line AA”therefore contains B and C in fixed proportion given by
[B : C]AA” = (β)(γ) : (1-α)(1-γ)
LIKEWISE BB” and CC” contain the fixed proportions
[C : A]BB” = (1-α)
(1-β)(δ)/(1-α-β+2αβ) : (1-β)(1-δ)
= (1-α)(δ)/(1-α-β+2αβ) : (1-δ)
[A : B]CC” =
(α)*(1-γ)(1-δ)/ (1-γ-δ+2γδ) : (α)(β)/(1-α-β+2αβ)*(γ)(δ)/ (1-γ-δ+2γδ)
= (1-γ)(1-δ) : (β)(γ)(δ)/(1-α-β+2αβ)
NOW lines AA”, BB” and CC” are concurrent precisely when we have
[C : A]BB” x [A : B]CC” =
(δ)/ (1-α-β+2αβ)*(1-γ) (1-δ) : (1-δ)* (β)(γ) (δ)/ (1-α-β+2αβ) =
(1-α)(1-γ) : (β)(γ) = 1/[B : C]AA”
In other words, along the new Cevian lines constructed through points A”, B” and C” the proportions of B : C, C : A and A : B commute just as they would along AA’, BB’ and CC’ and the reference triangle ABC is perspective with any Cevian triangle A”B”C” of its own Cevian triangle A’B’C’.
(This completes the proof )
PLEASE IGNORE (TEST ONLY):