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.

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!

(Click here for a larger image)

A TRIANGLE ** ABC** (above) contains a Cevian point

**with associated Cevian triangle**

*X**. In turn,*

**A’B’C’***contains a Cevian point*

**A’B’C’****with Cevian triangle (sides not shown)**

*Y***. If sides**

*A”B”C”***and**

*AB***are divided in the affine proportions β : (1-β) and α : (1-α) at**

*CA***and**

*C’***respectively if follows from Cevas Theorem that side**

*B’***is divided in affine proportion (αβ)/(1-α-β+2αβ) : (1-α)(1-β)/(1-α-β+2αβ) at**

*BC***. In other words since**

*A’**is Cevian to*

**A’B’C’**

**ABC**(α)*(β)*(1-α)*(1-β)/(1-α-β+2αβ) = (1-α)*(1-β)*(αβ)/(1-α-β+2αβ)

LIKEWISE sides **B’C’*** C’A’* and

*of triangle*

**A’B’***are divided in affine proportions*

**A’B’C’**** B’A”** :

**= γ : (1-γ)**

*A”C’***:**

*C’B”***= δ : (1-δ)**

*B”A’*IT follows

* A’C”* :

**= (1-γ)(1-δ)/(1 – γ -δ + 2γδ) : (cδ)/(1 – γ -δ + 2γδ)**

*C”B’*NOW the line ** AA”** is a barycentric combination of

**and**

*A***which necessarily contains**

*A”**and*

**B’***in fixed proportion. And whilst each vertex will contain proportions of both*

**C’***and*

**A***and*

**C***and*

**A***respectively, only the proportions of*

**B***(in*

**C***) and*

**B’***(in*

**B***) determine how*

**C’***will divide side*

**AA”***. Line*

**BC***therefore contains*

**AA”***and*

**B***in fixed proportion given by*

**C**[** B** :

**]**

*C***given by**

_{AA”}**(β)(γ) : (1-α)(1-γ)**

LIKEWISE ** BB”** and

**contain the fixed proportions**

*CC”*[** C** :

**]**

*A***= (1-α)**

_{BB”}= (1-α)(δ)/(1-α-β+2αβ) : (1-δ)

[** A** :

**]**

*B***=**

_{CC”}=** **(1-γ)(1-δ) : (β)(γ)(δ)/(1-α-β+2αβ)

NOW lines * AA”*,

*and*

**BB”***are concurrent precisely when we have*

**CC”**[** C** :

**]**

*A*

_{BB”}*x*[

**:**

*A***]**

*B*

_{CC” }

_{ = }(1-α)~~(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”*,

*and*

**B”***the proportions of*

**C”***:*

**B***,*

**C***:*

**C***and*

**A***:*

**A***commute just as they would along*

**B***,*

**AA’***and*

**BB’***and the reference triangle*

**CC’****is perspective with**

*ABC**any*Cevian triangle

**of its own Cevian triangle**

*A”B”C”***.**

*A’B’C’*(This completes the proof )