PlanetMath: proof of Van Aubel theorem

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proof of Van Aubel theorem

(Proof)

We want to prove

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

$\includegraphics{aubel}$

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

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

and

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

From the two equalities we can get

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

and thus

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

Since

, substituting leads to

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

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

and so

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

Subsituting the last equality gives the desired result.

"proof of Van Aubel theorem" is owned by drini. [ owner history (1) ]

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Other names:

Van Aubel's theorem

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Cross-references: Ceva's theorem, equalities, bisector's, angle

This is version 3 of proof of Van Aubel theorem, born on 2003-10-28, modified 2003-10-31.
Object id is 5413, canonical name is ProofOfVanAubelTheorem.
Accessed 3153 times total.

Classification:

51N20 (Geometry :: Analytic and descriptive geometry :: Euclidean analytic geometry)

Pending Errata and Addenda

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