Does a trapezoid cut into 2 parts form two similar trapezoids? Consider a trapezoid ABCD with AB less than CD, AB parallel to CD, right angled at A and D. If a segment EF is drawn from AD to BC parallel to the bases, are the two interior trapezoids similar?
Additionally, the intersection of the diagonals of trapezoid ABCD lies on the segment EF.
I know in this instance that all trapezoid angles must be equal, but I’m not sure if this enough to assume trapezoid similarity.
 A: Denote by $O$ the intersection of the diagonals. Then we have from the similar triangles
$$\frac{EO}{AB}=\frac{DO}{DB}$$ 
and 
$$\frac{EO}{CD}=\frac{AO}{AC} = \frac{BO}{DB}$$
so adding up the equalities we get
$$\frac{EO}{AB}+\frac{EO}{CD}= \frac{DO}{DB}+  \frac{BO}{DB}= 1$$
and similarly
$$\frac{FO}{AB}+\frac{FO}{CD}=1$$
Conclusion: $EO=FO$ and 
$$\frac{1}{EF}=\frac{1}{2}(\frac{1}{AB}+\frac{1}{CD})$$
so $EF$ is the harmonic mean of $AB$, $CD$ (and so not the geometric mean). 
$\bf{Added:}$ For a general parallel segment $EF$ its length is a weighted average of $AB$ and $CD$
$$EF = \frac{ED}{AD} AB + \frac{AE}{AD} CD$$
If you take the ratio
$\frac{AE}{ED} = \sqrt{\frac{AB}{CD}}$  you will get $EF = \sqrt{AB \cdot CD}$ and the two little trapezoids similar. Notice you need to draw $EF$ a bit lower than the intersection of the diagonal, but above the middle line. 
A: 
Since $AB\parallel EF\parallel CD$, then$$\frac{AE}{EC}=\frac{AG}{GD}=\frac{BF}{FD}$$and hence for corresponding sides in the two trapezoids$$\frac{AE}{EC}=\frac{BF}{FD}$$
But since by similar triangles$$\frac{AE}{EC}=\frac{AG}{GD}=\frac{AB}{CD}<\frac{EF}{CD}$$then the bases of the equiangular trapezoids are not proportional to their corresponding sides, and the trapezoids are not similar.
