Given triangle ABC, construct a line QR through B, such that QR is parallel to AC. The line AR intersects side BC at X and CQ intersects side AB at Z. Let P be the intersection of AR and CQ. Assume BP intersects side AC at Y. Thus, the Cevians AX, BY, and CZ are concurrent.[br][br]We know that QR is parallel to AC. So, by using Euclid's Proposition 29 to find congruent angles we can show that there are four pairs of similar triangles:[br]AZC~BZQ,[br]BXR~CXA,[br]CYP~QBP,[br]YAP~BRP.[br]From the similarities, we can get equal ratios:[br][math]\frac{AZ}{BZ}=\frac{AC}{BQ}[/math],[br][math]\frac{BX}{CX}=\frac{BR}{CA}[/math],[br][math]\frac{CY}{QB}=\frac{PY}{PB}[/math],[br][math]\frac{YA}{BR}=\frac{YP}{BP}.[/math][br]From the last two equations, we obtain the equal ratios [math]\frac{CY}{YA}=\frac{QB}{BR}[/math]. [br][br]Notice that [math]\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=\frac{AC}{BQ}\cdot\frac{BR}{CA}\cdot\frac{QB}{BR}=1.[/math]

The contrapositive of the converse of Ceva's Theorem says: [br]If AX, BY, and CZ are not concurrent, then [math]\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}\ne1.[/math][br][br]Given the triangle construction described in the proof of Ceva's theorem above, assume that AX, BY, and CZ are not concurrent. [br][br]Let AX and CZ intersect at a point P. BY does not intersect AX and CZ at P, but construct a new cevian BW such that AX,CZ,BW are concurrent. Then [math]\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CW}{WA}=1.[/math][br][br]Notice [math]\frac{CW}{WA}\ne\frac{CY}{YA}[/math], thus [math]\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}\ne\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CW}{WA}\ne1.[/math]