# Ceva's Theorem

Ceva's Theorem is an algebraic statement regarding the lengths of cevians in a triangle.

## Statement

(awaiting image)
A necessary and sufficient condition for AD, BE, CF, where D, E, and F are points of the respective side lines BC, CA, AB of a triangle ABC, to be concurrent is that $BD * CE * AF = +DC * EA * FB$

where all segments in the formula are directed segments.

## Proof

Let ${X,Y,Z}$ be points on ${BC}, {CA}, {AB}$ respectively such that $AX,BY,CZ$ are concurrent, and let ${P}$ be the point where $AX$, $BY$ and $CZ$ meet. Draw a parallel to $AB$ through the point ${C}$. Extend $AX$ until it intersects the parallel at a point $\displaystyle{A'}$. Construct $\displaystyle{B'}$ in a similar way extending $BY$.

(ceva1.png)

The triangles ${\triangle{ABX}}$ and ${\triangle{A'CX}}$ are similar, and so are $\triangle{ABY}$ and $\triangle{CB'Y}$. Then the following equalities hold: $\begin{displaymath}\frac{BX}{XC}=\frac{AB}{CA'},\qquad\frac{CY}{YA}=\frac{CB'}{BA}\end{displaymath}$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

and thus $\begin{displaymath} \frac{BX}{XC}\cdot\frac{CY}{YA}=\frac{AB}{CA'}\cdot\frac{CB'}{BA}=\frac{CB'}{A'C}. \end{displaymath} (1)$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

Notice that if directed segments are being used then $AB$ and $BA$ have opposite signs, and therefore when cancelled change the sign of the expression. That's why we changed $CA'$ to $A'C$.

Now we turn to consider the following similarities: $\triangle{AZP}\sim\triangle{A'CP}$ and $\triangle BZP\sim\triangle B'CP$. From them we get the equalities $\begin{displaymath}\frac{CP}{ZP}=\frac{A'C}{AZ},\qquad\frac{CP}{ZP}=\frac{CB'}{ZB}\end{displaymath}$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

which lead to $\begin{displaymath}\frac{AZ}{ZB}=\frac{A'C}{CB'}.\end{displaymath}$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

Multiplying the last expression with (1) gives $\begin{displaymath}\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1\end{displaymath}$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

and we conclude the proof.

To prove the converse, suppose that $X,Y,Z$ are points on ${BC, CA, AB}$ respectively and satisfying $\begin{displaymath}\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1.\end{displaymath}$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

Let $Q$ be the intersection point of $AX$ with $BY$, and let $Z'$ be the intersection of $CQ$ with $AB$. Since then $AX,BY,CZ'$ are concurrent, we have $\begin{displaymath}\frac{AZ'}{Z'B}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1\end{displaymath}$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

and thus $\begin{displaymath}\frac{AZ'}{Z'B}=\frac{AZ}{ZB}\end{displaymath}$ (Error compiling LaTeX. ! LaTeX Error: Bad math environment delimiter.)

which implies $Z=Z'$, and therefore $AX,BY,CZ$ are concurrent.

## Example

Suppose AB, AC, and BC have lengths 13, 14, and 15. If $\frac{AF}{FB} = \frac{2}{5}$ and $\frac{CE}{EA} = \frac{5}{8}$. Find BD and DC.

If $BD = x$ and $DC = y$, then $10x = 40y$, and ${x + y = 15}$. From this, we find $x = 12$ and $y = 3$.