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 $B C$ , $C A$ and $A B$ , respectively, such that $A X$ , $B Y$ and $C Z$ are concurrent, and let $P$ be the point where $A X$ , $B Y$ and $C Z$ meet.Draw a parallel to $A B$ through the point $C$ . Extend $A X$ until it intersects the parallel at a point $A^{\\prime}$ . Construct $B^{\\prime}$ in a similar way extending $B Y$ .

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 $A B$ and $B A$ 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}

which lead to

\\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 $B C, C A, A B$ respectively and satisfying

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

Let $Q$ be the intersection point of $A X$ with $B Y$ , and let $Z^{\\prime}$ be the intersection of $C Q$ with $A B$ . 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 $A X, B Y, C Z$ are concurrent.