<Tri-2024-2109>, hint

Let M be the midpoint of arc BC, then A,I,M,E are collinear and M is the center of circle BIC 

<BIM=<IBM. Le's prove IJ/IE=AD/AE. IJ/IE= r/2R sin(A/2)=

4R sin(A/2) sin(B/2) sin(C/2)/ 2R sin(A/2)=2 sin(B/2) sin(C/2)

AD/AE=2R sinC sinB/(AM+ME)(=2Rsin(B+A/2)+2R sin (A/2)=2sin(B/2) sin(C/2)

sin X+sinY=2sin(X+Y/2) cos(X-Y/2) Hence Triangles AED and IEJ are similar

Hence E,J,D are collinear