# proof of Ceva’s theorem

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As in the article on Ceva’s Theorem, we will consider directed line segments.

Let $X$, $Y$ and $Z$ be points on $BC$, $CA$ and $AB$, respectively, such that $AX$, $BY$ and $CZ$ 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 $A^{\prime}$. Construct $B^{\prime}$ in a similar way extending $BY$.

The triangles $\triangle ABX$ and $\triangle A^{\prime}CX$ are similar, and so are $\triangle ABY$ and $\triangle CB^{\prime}Y$. Then the following equalities hold:

 $\frac{BX}{XC}=\frac{AB}{CA^{\prime}},\qquad\frac{CY}{YA}=\frac{CB^{\prime}}{BA}$

and thus

 $\frac{BX}{XC}\cdot\frac{CY}{YA}=\frac{AB}{CA^{\prime}}\cdot\frac{CB^{\prime}}{% BA}=\frac{CB^{\prime}}{A^{\prime}C}.$ (1)

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 $CA^{\prime}$ to $A^{\prime}C$.

Now we turn to consider the following similarities: $\triangle AZP\sim\triangle A^{\prime}CP$ and $\triangle BZP\sim\triangle B^{\prime}CP$. From them we get the equalities

 $\frac{CP}{ZP}=\frac{A^{\prime}C}{AZ},\qquad\frac{CP}{ZP}=\frac{CB^{\prime}}{ZB}$

 $\frac{AZ}{ZB}=\frac{A^{\prime}C}{CB^{\prime}}.$

Multiplying the last expression with (1) gives

 $\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1$

and we conclude the proof.

To prove the converse, suppose that $X,Y,Z$ are points on $BC,CA,AB$ respectively and satisfying

 $\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1.$

Let $Q$ be the intersection point of $AX$ with $BY$, and let $Z^{\prime}$ be the intersection of $CQ$ with $AB$. Since then $AX,BY,CZ^{\prime}$ are concurrent, we have

 $\frac{AZ^{\prime}}{Z^{\prime}B}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1$

and thus

 $\frac{AZ^{\prime}}{Z^{\prime}B}=\frac{AZ}{ZB}$

which implies $Z=Z^{\prime}$, and therefore $AX,BY,CZ$ are concurrent.

Title proof of Ceva’s theorem ProofOfCevasTheorem 2013-03-22 12:38:54 2013-03-22 12:38:54 yark (2760) yark (2760) 11 yark (2760) Proof msc 51A05