I found this problem online, where it was asked to prove EF = GH.
I was able to prove that, but got intrigued by how the four smaller inscribed circles could be constructed in the first place.
That is, given the fact that AD + BC = AB + CD (i.e. an inscribed circle can be constructed for ABCD), how can we contruct EF and GH, such that for each smaller quadrilateral, an incribed circle can be constructed?
Also, given a fixed quadrilateral with AD + BC = AB + CD, are the positions of EF and GH unique?
Any hints would be appreciated.


Hint:
Use this fact:
ُThe center of four circle I, J, K and L are on a circle (M) concentric with inscribed circle N at O and Lines AO, BO, CO and DO which connect vertices of quadrilateral to the center (O) of inscribed circle.
Now follow this rule:
1- Draw AO. BO, CO and DO.
2-draw AC and BD, they intersect on P.
3-Connect OP.
4- find the midpoint of OP and mark it as Q. Q is the intersection of EF and GH (in your figure P).
5- Draw a line parallel with BD, it intersect BO and DO on J and L respectively.these are the center of two opposite circles.They also define the measure of the radius (M).
6- Draw circle (M), it intersect AO and CO on I and K respectively. these are the centers of other two opposite circles.
7- Draw four cicles.
8- draw common tangents of adjacent circles, they will intersect on Q( in your figure P) and you get EF and GH.
The reason for taking Q as the midpoint of OP is that if equilateral is regular, points P and O and Q will be coincident. If it is not regular , they will have a distance like OP and the intersection of common tangents locates on its midpoint Q.