# proof of trigonometric version of Ceva’s theorem

The proof goes by proving the condition imposed on the sines, is equivalent to the one imposed on the sides on the normal version of Ceva’s theorem.

We want to prove

 $\frac{\sin ACZ}{\sin ZCB}\cdot\frac{\sin BAX}{\sin XAC}\cdot\frac{\sin CBY}{% \sin YBA}=1$

if and only if

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

Now, the generalization of bisectors theorem states that

 $\displaystyle\frac{AZ}{ZB}$ $\displaystyle=\frac{CA\sin ACZ}{BC\sin ZCB},$ $\displaystyle\frac{BX}{XC}$ $\displaystyle=\frac{AB\sin BAX}{CA\sin XAC},$ $\displaystyle\frac{CY}{YA}$ $\displaystyle=\frac{BC\sin CBY}{AB\sin YBA}.$

Multiplying the three equations, and cancelling all segments on the right side gives the desired equivalence.

Title proof of trigonometric version of Ceva’s theorem ProofOfTrigonometricVersionOfCevasTheorem 2013-03-22 14:49:20 2013-03-22 14:49:20 drini (3) drini (3) 7 drini (3) Proof msc 51A05