Prove that the dual (ie, "midpoint polygon") of an isosceles trapezoid is a rhombus. Prove that the dual of an isosceles trapezoid is a rhombus.


Here, the "dual" of any polygon is where its sides intersects the midpoint of each side of the "outer" figure. (In other words, it's the "midpoint polygon".)

I'm aware of the properties of an isosceles trapezoid:


By definition, the legs are congruent.
The lower and upper base angles are congruent.
Any lower base angle is supplementary to any upper base angle.


Here is my drawing of the dual polygon in the isosceles trapezoid: 



I showed that $\triangle FAE\cong\triangle HDE$ by SAS, and that $\triangle GBF\cong\triangle GCH$ by SAS. So by CPCTC, $\overline{FE}\cong\overline{HE}$, and $\overline{FG}\cong\overline{GH}$.


  I'm unsure of how to show that all of the sides of the dual quadrilateral are congruent, making it a rhombus.
By properties of the midpoint polygon, each side of the midpoint polygon is parallel and equal to half the length of a diagonal (why?). An isosceles trapezoid has diagonals of equal length, hence the result. 


   Consider the triangle $\Delta ABC$. As $AF=FB$ and $BG=GC$, we get $\Delta BFG\sim\Delta BAC$ and the similarity ratio is a half. The other sides of the midpoint polygon follows somewhat similarly. 


Compare with Varignon's theorem.
Since G and E are the base midpoints of the isosceles trapezoid, BC $\perp$ GE. Since  F and H are the side midpoints of the isosceles trapezoid, BC || FH. Therefore, GE $\perp$ FH. 

Since GOHC and GEDC are similar, GO:OE = CH:HD = 1:1. Also, FO = OH = $\frac14$ (BC + AD). Therefore, O bisects the perpendicular GE and FH, which leads to that the four right triangles are congruent, hence a rhombus.