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 $\mathrm{△}ABX$ and $\mathrm{△}{A}^{\prime }CX$ are similar, and so are $\mathrm{△}ABY$ and $\mathrm{△}C{B}^{\prime }Y$. Then the following equalities hold:

$$\frac{BX}{XC}=\frac{AB}{C{A}^{\prime }},\frac{CY}{YA}=\frac{C{B}^{\prime }}{BA}$$

and thus

$$\frac{BX}{XC}\cdot \frac{CY}{YA}=\frac{AB}{C{A}^{\prime }}\cdot \frac{C{B}^{\prime }}{BA}=\frac{C{B}^{\prime }}{A^{\prime }C}.$$ (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 $C{A}^{\prime }$ to $A^{\prime }C$.

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

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

which lead to

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

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 $AX,BY,C{Z}^{\prime }$ are concurrent, we have

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

and thus

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

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

Title	proof of Ceva’s theorem
Canonical name	ProofOfCevasTheorem
Date of creation	2013-03-22 12:38:54
Last modified on	2013-03-22 12:38:54
Owner	yark (2760)
Last modified by	yark (2760)
Numerical id	11
Author	yark (2760)
Entry type	Proof
Classification	msc 51A05