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For an arbitrary convex quadrialteral $ABCD$:

$$AF=FC, BE=ED, GI=IH$$

Prove that points $F$, $E$ and $I$ are collinear.

I was able to solve the problem by using vectors and I think that it can be also solved analytically but I wanted some more elegant proof, more in the spirit of Euclid.

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  • $\begingroup$ Does a solution using projective transformations on line counts? $\endgroup$
    – nonuser
    Dec 26, 2017 at 16:57
  • $\begingroup$ Yes, it counts! $\endgroup$
    – Saša
    Dec 26, 2017 at 17:05
  • $\begingroup$ The midpoint of $AC$ is not the only point fulfilling $AF=FC$, better to state the constraints as "$F$ is the midpoint of $AC$" and so on. Related: mathworld.wolfram.com/CompleteQuadrilateral.html $\endgroup$ Dec 26, 2017 at 17:15

2 Answers 2

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Quoting Johnson, Advanced Euclidean Geometry:

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A projective solution:

Fix points $A$, $B$, $C$ and $G$ and move $D$ on (fixed) line $AG$, so that $H$ moves on (fixed) line $AB$. Point $E$ moves on parallel (middle line) $p$ to $AG$ and point $I$ moves on parallel (middle line) $q$ to $AB$. (Let $p\cap q=\{S\}$, it is easy to see that $S$ halves $BG$!) So $$E\mapsto D \mapsto H\mapsto I$$ is projective transformation from line $p$ to line $q$ since it is composite of $3$ perspective transformation respectively with centers at $B$, $C$ and $G$. Now this transformation is also perspective with respect to some point $T$ since when $E=S$ then $D=G$ and $I=S$, so $S$ maps to it self.

We are left to identify point $T$. And it is not difficult to see that $T=E$.

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