# proof of Van Aubel theorem

We want to prove

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

On the picture, let us call $\phi$ to the angle $\angle ABE$ and $\psi$ to the angle $\angle EBC$.

A generalization of bisector’s theorem states

 $\frac{CE}{EA}=\frac{CB\sin\psi}{AB\sin\phi}\quad\mbox{on }\triangle ABC$

and

 $\frac{CP}{PF}=\frac{CB\sin\psi}{FB\sin\phi}\quad\mbox{on }\triangle FBC.$

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

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

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 ProofOfVanAubelTheorem 2013-03-22 14:03:28 2013-03-22 14:03:28 drini (3) drini (3) 6 drini (3) Proof msc 51N20 Van Aubel’s theorem ProofOfVanAubelsTheorem CevasTheorem VanAubelTheorem TrigonometricVersionOfCevasTheorem