We will use the notation ([math]\Delta[/math]XYZ) to denote the area of an arbitrary triangle [math]\Delta[/math]XYZ.[br][br]Ceva's Theorem: In triangle [math]\Delta[/math]ABC, if the Cevians AX, BY, and CZ are concurrent, then [math]\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1[/math]. [br][br]Proof: Let [math]\Delta[/math]ABC be a triangle with Cevians AX, BY, and CZ being concurrent. Let BD, AE, and CF be the altitudes of ABC. Let OG be an altitude of [math]\Delta[/math]AOC, OH be an altitude of [math]\Delta[/math]COB, and OI be an altitude of [math]\Delta[/math]BOA. Notice the following three things:[br][br]1) [math]\left(\frac{\left(ABY\right)}{\left(CBY\right)}\right)=\frac{\frac{1}{2}\left(AY\right)\left(BY\right)}{\frac{1}{2}\left(CY\right)\left(BY\right)}=\frac{AY}{CY}=\frac{\frac{1}{2}\left(AG\right)\left(BG\right)}{\frac{1}{2}\left(CG\right)\left(BG\right)}=\left(\frac{\left(AOG\right)}{\left(COG\right)}\right)[/math][br][br]2) [math]\left(\frac{\left(CAZ\right)}{\left(BAZ\right)}\right)=\frac{\frac{1}{2}\left(CZ\right)\left(AZ\right)}{\frac{1}{2}\left(BZ\right)\left(AZ\right)}=\frac{CZ}{BZ}=\frac{\frac{1}{2}\left(CH\right)\left(OH\right)}{\frac{1}{2}\left(BH\right)\left(OH\right)}=\left(\frac{\left(AOH\right)}{\left(BOH\right)}\right)[/math][br][br]3) [math]\left(\frac{\left(BCX\right)}{\left(ACX\right)}\right)=\frac{\frac{1}{2}\left(BX\right)\left(CX\right)}{\frac{1}{2}\left(AX\right)\left(CX\right)}=\frac{BX}{AX}=\frac{\frac{1}{2}\left(BI\right)\left(OX\right)}{\frac{1}{2}\left(AI\right)\left(OX\right)}=\left(\frac{\left(BOX\right)}{\left(AOX\right)}\right)[/math][br][br][br]Now by multiplying the ratios together, we see that [math]\frac{AY}{CY}\cdot\frac{CZ}{BZ}\cdot\frac{BX}{AX}=\frac{ABCXYZ}{ABCXYZ}=1[/math]. [br][br]Therefore, in triangle [math]\Delta[/math]ABC, if the Cevians AX, BY, and CZ are concurrent, then [math]\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1[/math].

Converse of Ceva's Theorem: In triangle ABC, if the product of [math]\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1[/math], then the Cevians AX, BY, and CZ are concurrent. [br][br][br]Proof: Let ABC be a triangle with [math]\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1[/math]. We wish to show that AX, BY, and CZ are concurrent. Assume that Cevians AZ and CX intersect at O, and that the other Cevian through O is BH. By Ceva's Theorem, we have that [math]\frac{AH}{HB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1[/math]. However, by our given we know that [math]\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=1[/math]. Thus, we can see that [math]\frac{AH}{HB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}=\frac{AZ}{ZB}\cdot\frac{BX}{XC}\cdot\frac{CY}{YA}[/math]. From this, we can imply that [math]\frac{AH}{HB}=\frac{AZ}{ZB}[/math]. Note that this can only be true if H and Z are the same point. Therefore, we can conclude that AX, BY, and CZ are concurrent because we assumed that AZ, CX, and BH were concurrent.