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
Canonical name	ProofOfTrigonometricVersionOfCevasTheorem
Date of creation	2013-03-22 14:49:20
Last modified on	2013-03-22 14:49:20
Owner	drini (3)
Last modified by	drini (3)
Numerical id	7
Author	drini (3)
Entry type	Proof
Classification	msc 51A05