proof of Van Aubel theorem

We want to prove

$$\frac{CP}{PF}=\frac{CD}{DB}+\frac{CE}{EA}$$

On the picture, let us call $\varphi$ to the angle $\mathrm{\angle }ABE$ and $\psi$ to the angle $\mathrm{\angle }EBC$.

A generalization of bisector’s theorem states

$$\frac{CE}{EA}=\frac{CB\sin \psi }{AB\sin \varphi }\mathit{\hspace{1em}}\text{on }\mathrm{△}ABC$$

and

$$\frac{CP}{PF}=\frac{CB\sin \psi }{FB\sin \varphi }\mathit{\hspace{1em}}\text{on }\mathrm{△}FBC.$$

From the two equalities we can get

$$\frac{CE\cdot AB}{EA}=\frac{CP\cdot FB}{PF}$$

and thus

$$\frac{CP}{PF}=\frac{CE\cdot AB}{EA\cdot FB}.$$

Since $AB=AF+FB$, substituting leads to

${\displaystyle \frac{CE\cdot AB}{EA\cdot FB}}$ $={\displaystyle \frac{CE(AF+FB)}{EA\cdot FB}}$

$={\displaystyle \frac{CE\cdot AF}{EA\cdot FB}}+{\displaystyle \frac{CE\cdot FB}{EA\cdot FB}}$

$={\displaystyle \frac{CE\cdot AF}{EA\cdot FB}}+{\displaystyle \frac{CE}{EA}}$

But Ceva’s theorem states

$$\frac{CE}{EA}\cdot \frac{AF}{FB}\cdot \frac{BD}{DC}=1$$

and so

$$\frac{CE\cdot AF}{EA\cdot FB}=\frac{CD}{DB}$$

Subsituting the last equality gives the desired result.

Title	proof of Van Aubel theorem
Canonical name	ProofOfVanAubelTheorem
Date of creation	2013-03-22 14:03:28
Last modified on	2013-03-22 14:03:28
Owner	drini (3)
Last modified by	drini (3)
Numerical id	6
Author	drini (3)
Entry type	Proof
Classification	msc 51N20
Synonym	Van Aubel’s theorem
Related topic	ProofOfVanAubelsTheorem
Related topic	CevasTheorem
Related topic	VanAubelTheorem
Related topic	TrigonometricVersionOfCevasTheorem