In the figure there is a convex quadrilateral ABCD. The lines DA and CB intersect at K, the lines AB and DC intersect at L, the lines AC and KL intersect at G, the
lines DB and KL intersect at F. Prove that $\frac{KF}{FL} = \frac{KG}{GL}$ .
I feel like this question is screaming for menelaus but i can't figure out where; so i did what any desperate person would do in this case; write out all the menelaus for all of the collinear triples. But there were 10 triples so all I got was a page full of letters from which I couldn't figure out anything.
I would highly appreciate a synthetic solution using things like Menelaus, Ceva's, Area Lemma and just basic things (nothing too fancy like complex or bary)
Thanks!
Best Answer
If you use Menelaus, you obtain $$\frac{LG \cdot KA \cdot DC}{GK \cdot AD \cdot CL} = 1,$$ where we ignore the signs of lengths.
Similarly, Ceva yields $$\frac{LF \cdot AK \cdot DC}{FK \cdot AD \cdot CL} = 1.$$
It follows that $\frac{LG}{GK} = \frac{LF}{FK}.$
I ignored signed length out of my haste, but this is the general method of showing the result. As mentioned by Aqua, this used frequently in projective geometry, specifically because the four points form a harmonic bundle. I urge you to search that up, if you are interested :)