Similarity conditions of two right trapezoid with similar angles We have $2$ right trapezoid for example two trapezoid with angles $90^{\circ},90^{\circ},80^{\circ},100^{\circ}$.
do we need to all the sides proportionality or less is enough ?
 A: The trapeziums are not drawn with angles you mentioned.. so as to bring it into more contrast. As seen in the diagram the trapeziums $ABCD, ABEF $ are not similar.

Two trapezoids with the same angles are similar only if the ratio of corresponding sides in those similar triangles is the same after these corresponding sides are parallelly placed to appear to radiate from a single center point of similitude $O$.
A: Two trapezoids with the same angles are similar if and only the ratio between one particular pair of neighboring sides is the same in two trapezoids.
If the trapezoid is not a parallelogram, it is also enough to know that the ratio between the two parallel sides is the same in the two trapezoids.
A: If the ratios between the bases are the same, say $k$, the trapezoids are similar.
Indeed, both trapezoids $ABCD$ and $A'B'C'D'$ can be consedered as "pieces" of right triangles with acute angles $80$ and $10$ degrees. The triangles have been cut along a segment parallel to the side opposite to the $10$ degrees angle.
Let $AB$ and $CD$ the parallel sides, $AD$ the perpendicular side and $BC$ the crosswise side.  The same for $A'B'C'D'$. Say that
$$\frac{AB}{CD}=\frac{A'B'}{C'D'}=k$$
Suppose that they are pieces of triangles $ECD$ and $E'C'D'$. Since the trapezoids have their angles equal, to show that both trapezoids are similar it suffices to check if the ratios between corresponding sides are the same. For example:
$$\frac{AD}{A'D'}=\frac{ED-EA}{E'D'-E'A'}=\frac{ED-k\cdot ED}{E'D'-k\cdot E'D'}=\frac{ED}{E'D'}$$
where $EA=k\cdot ED$ because the triangles $EAB$ and $ECD$ are similar.
