Isogonal Concurrencies

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25 February 2015, Created with GeoGebra

Acknowledgment

The applet above illustrates four distinct concurrencies associated with the triangle formed by isogonally conjugate cevians. The first three have been dealt with early on at Morley's triangle discussion and a separate page. The fourth one has been brought to my attention by Daniel Hardisky who posted the result and a proof at the CutTheKnotMath facebook page.

Below I give a simplified version of the proof.

Problem

Let in $\Delta ABC$ the pairs of lines $AE,AG;$ $BE,BF;$ $CF,CG$ be isogonal conjugate with respect to angles $BAC, CBA, ACB.$ Let $A',B',C'$ be the projections of $F,G,E$ on the sides $BC,AC,AB,$ respectively.

isogonal concurrency, problem

Then lines $AA',BB',CC'$ are concurrent.

Proof

The proof is elementary and depends on the inverse of Ceva's Theorem.

isogonal concurrency, solution

Let $\angle C'AE=\angle B'AG=\alpha,$ $\angle C'BE=\angle A'BF=\beta,$ $\angle A'CF=\angle B'CG=\gamma,$ $FA'=m.$ Then $BA'=m\cdot\cot\beta,$ $CA'=m\cdot\cot\gamma$ so that

$\displaystyle\frac{CA'}{BA'}=\frac{\cot\beta}{\cot\gamma}.$

Similarly,

$\begin{align}\displaystyle \frac{CB'}{AB'} &=\frac{\cot\gamma}{\cot\alpha},\\ \frac{AC'}{BC'} &=\frac{\cot\alpha}{\cot\beta} \end{align}$

Multiplying the three yields Ceva's condition $\displaystyle\frac{CA'}{BA'}\cdot\frac{CB'}{AB'}\cdot\frac{AC'}{BC'}=1$ which exactly says that $AA',BB'CC'$ are concurrent.

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