# Desargues' Hexagon

The GeoGebra applet below illustrates an observation by Vladimir Nikolin (June, 2012).

Let $ABC$ and $DEF$ be two triangles in perspective, meaning, for example, that lines $AE,$ $BF,$ and $CD$ are concurrent. Then points $G = AB\cap DF,$ $H = AC\cap DF,$ $I = AC\cap EF,$ $J = BC\cap EF,$ $K = BC\cap DE,$ and $L = AB\cap DE$ lie on a conic.

Points $G,$ $H,$ $I,$ $J,$ $K,$ $L$ are pairwise intersections of the sides of the two given triangles.

The applet displays two triangles that initially are not perspective. To achieve that goal, drag either one of the six vertices or the triangles as a whole. The applet displays a conic through five points $G,$ $H,$ $I,$ $J,$ and $K.$ Observe that when the three lines $AE,$ $BF,$ and $CD$ are made concurrent, point $L$ falls on the same conic.

17 January 2015, Created with GeoGebra

The proof is short. By Desargues' theorem, triangles persepctive in a point are perspective in a line. The converse of Pascal's theorem tells us that the hexagon $ABCDEF$ is inscribed in a conic. Vladimir's observation is that the same is true of the hexagon $IJKLGH.$ This is so because the sides of the latter lie on the sides of the two triangles, for example, $GL$ lies on $AB,$ $GH$ on $DF,$ etc. As a consequence, the three intersections of the opposite sides of $IJKLGH$ are collinear, allowing for another application of Pascal's theorem converse.

I think that "Desargues' Hexagon" is a proper moniker for the hexagon $IJKLGH.$

### Desargues' Theorem

- Desargues' Theorem
- 2N-Wing Butterfly Problem
- Cevian Triangle
- Do You Speak Mathematics?
- Desargues in the Bride's Chair (with Pythagoras)
- Menelaus From Ceva
- Monge from Desargues
- Monge via Desargues
- Nobbs' Points, Gergonne Line
- Soddy Circles and David Eppstein's Centers
- Pascal Lines: Steiner and Kirkman Theorems II
- Pole and Polar with Respect to a Triangle
- Desargues' Hexagon
- The Lepidoptera of the Circles

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