The obvious conjecture is that the midpoint is fixed as X is moved anywhere in the plane.
For one approach to a proof, consider the line through A and B and the perpendiculars to AB through X, Y, and Z.
Examine the triangles indicated for congruencies. See what happens for various locations of X.
Click here for the GSP file giving the above figure. Here is the basic outline of a proof.
1. AY and AX are the same length
2. Angle XAI and angle IAY are complementary.
3. Therefore angle AXI is congruent to angleYAI; angle AYK is congruent to angle XAI.
4. Triangle XAI is congruent to triangle AYK by ASA.
5. Similarly, triangle XBI is congrunent to triangle BZJ.
6. Consider trapezoid. KYZJ. The sum of the lengths of KY and JZ is equal to the length of AB.
7. For different locations of X, the lengths of KY and JZ may vary but AB is a fixed length that does nor vary.
8. The length of LD is the arithmetic mean of the lengths of KY and JZ.
9. Therefore the length of LD is one-half the length of AB.
10. Hence point D is in a fixed location with respect to AB and that location is indpendent of the location of X.
Note: Still to be considered: Locations of X that are such that KYZJ is not a trapezoid. The directed distances of AI and BI will have to be taken into account.