Let $BCED$ be a quadrilateral.
- $BE$ and $DC$ intersect at $G$,
- $BD$ and $CE$ intersect at $A$,
- $DE$ and $BC$ intersect at $F$,
- $AG$ and $BC$ intersect at $H$,
- $FG$ and $CE$ intersect at $I$,
- $AF$ and $BE$ intersect ta $J$.
We need to prove that $H, I$ and $J$ are collinear. I can prove it using projective geometry:
My solution: Triangle $AGF$ and $BCE$ are perspective from the point $D$, and hence by Desargues' theorem, they are perspective from a line, i.e the points of intersection of their corresponding sides (w.r.t this perspectivity) will be collinear. Now from intersection 4, 5 and 6, we see that $H, I, J$ are indeed collinear.
But I rather want a solution in Euclidean geometry.
Best Answer
Using Menelaus's theorem,
$\frac{FJ}{JA}$$\frac{AH}{HG}$$\frac{GI}{IF}$
= $\frac{FE}{ED}$$\frac{DB}{BA}$ $\frac{AH}{HG}$ $\frac{GA}{AH}$$\frac{HC}{CF}$
= $\frac{EF}{ED}$$\frac{DB}{BA}$ $\frac{GA}{HG}$ $\frac{HC}{CF}$
= $\frac{EF}{ED}$$\frac{DB}{BA}$ $\frac{AD}{DB}$$\frac{BC}{CH}$ $\frac{HC}{CF}$
= $\frac{EF}{ED}$$\frac{AD}{BA}$$\frac{BC}{CF}$
= 1
Therefore, H, I and J are collinear.