Problems about angle bisectors of a trapezium There is this geometry problem that I came across but need these conclusions in order to proceed. I really appreciate any helps. 
Let $ABCD$ be a trapezium with $AB$ and $DC$ parallel. The angle bisectors of $\angle A$ and $\angle D$ intersect at $E$, and the angle bisectors of $B$ and $C$ intersect at $F$.
I am trying to divide into cases what will happen if $E$ and $F$ overlap. 
If $E$ and $F$ overlap, then is it true that $AD+BC=AB+DC$? If yes, why?
If $E$ and $F$ do not overlap, then is it true that $EF$ parallel to both $AB$ and $DC$?
Are there any such theorems or propositions? I really appreciate any helps. 
Many many thanks.
 A: Since FB and FC are angle bisectors of $\angle B$ and $\angle C$ respectively, a semi-inscribed circle can be drawn with center at F such that FP, FQ and FR are its radius r. Refer to the first figure below. 

First of all, we need to show that [[PFR is in fact a straight line and hence the diameter of the circle.] 
Suppose it is not but PF (produced to cut DC at S) is. By alternate angles between parallel lines $\angle PSR$ is also a right angle. In this case, we have a triangle with two of its angles being right angle. This contradicts our assumption. Thus, [[PFR is in fact a straight line.] And it is not difficult to see that it is also a diameter of the circle.]
Similarly, another circle GHK can be drawn with center at E such that EG, EH and EK are its radius.
HEK is a straight line and in fact in the diameter of the circle GHK. See the second figure.

Note that PS(or PR) and HK are perpendicular distances between the parallel lines AB and DC. Thus, PS = HK. Therefore, E and F are both at a distance r from AB. Hence, EF//AB.
In the case E coincides with F, we have just one circle. The said sum can be obtained by applying tangent properties.
