carlosbr wrote: Let $ABCD$ be a convex quadrilateral, let $E$ and $F$ be the midpoints of the sides $AD$ and $BO$, respectively. The segment $CE$ meets $DF$ in $O$. Show that if the lines $AO$ and $BO$ divide the side $CD$ in 3 equal parts, then $ABCD$ is a parallelogram. Hi, are you sure F is the midpoint of BO as you wrote?..i was just thinking F might be the midpoint of BC instead, i'm not sure i haven't tried the problem yet, but i was jus reading it

F is midpoint of BC $Tipe$

Let $T$ the intersection beetween $BO$ and $DF$. Now we know that $2[TDB]=[TBC]=2[TFB]$ and so $[TDB]=[TFB]$. But this implies $DO=OF$. Similary $OE=OC$ and so $DEFC$ is a parallelogramm because the diagonals bisects each other. But also $ABCD$ is a parallelogramm because $DC//EF//AB$.

isn't T = O?

Simo_the_Wolf wrote: Let $ T$ the intersection beetween $ BO$ and $ DF$. Now we know that $ 2[TDB] = [TBC]$ How do we know this? And yes, I think T is O Anyways, we can also use Menelaus to show OD=OF and OE=OC, again showing DEFC is a parallelogram and thus so is ABCD.

I still don't quite how you can prove synthetically that $DO = OF$.

DO = OF due to OT is the midline of the triangle DFS, because T is midpoint of DS and OT $ \parallel $ FS When T is BO $ \cap $ DC and S is AO $ \cap $ DC Same strategy we can prove that EO = OC, so DEFC is parallelogram...