Guys I need help and the internet is failing me.
I need to prove Wittenbauer's Parallelogram. This is what it looks like: Wittenbauer's Parallelogram from Interactive Mathematics Miscellany and Puzzles
Basically, any convex quadrilateral ABCD, if you trisect it's sides at S,T; U,V; W,X; Y,Z then TU, VW, XY, ZS form a parallelogram.
All I really have to work with is area and similarity and I'm just completely stumped. I've been staring at it for a while and just am not seeing what initial connection I have to make in order to see the proof.
Once I prove it's a parallelogram I have other stuff to prove and am hoping they follow easily afterward. Here is the other stuff:
Prove that |TU| = 2/3 |AC| and |VX| = 2/3 |BD|
If ABCD is a rhombus, prove that the figure formed by TU, VW, XY, ZS is a rectangle and that it's area is equal to 8/9 |ABCD|.
Re: Wittenbauer's Parallelogram
But this same page explains that the lines drawn are parallel to the diagonals of the given quadrilateral. Since there are only two diagonals, the lines come in parallel pairs and thus form a parallelogram.