About concylic points

Let the intersecting point of AG and EH be M.
For triangle EGM and triangle EMA,
EM=EM (common)..........(1)
Since CE = EA and BH = HA ( E and H are mid-point of CA and AB), therefore, by mid-point theorem, BC//EH. and GM = AM.........(2)
Also, angle EMG = angle EMA = 90 degree.........(3)
From (1), (2) and (3), triangle EGM congruent triangle EMA (SAS)
Therefore, angle GEM = angle AEM....................(4)
Again by mid-point theorem, DH//AC, therefore angle DHE = angle AEM (alternate angle DH//AC).....................................................(5)
Again by mid-point theorem, EH//BC, therefore, angle HEG = angle EGC (alternate angle EH//BC).............................................(6)
From results of (4),(5) and (6), we get angle DHE = angle EGC,
therefore, DGEH is a cyclic quadrilateral because exterior angle equal.
Therefore, D,G,E,H are concyclic.