<Tri-2024-2154>, hint

Let  line BE cut circumcircle  of ACE in F.

Since triangles ECD and CBD are similar CD^2=DE x DB

Since AECF are  cyclic,  AD xDC=DF xDE=CD^2. Hence DB=DF DA=DC  ABCF is a parallelogram

<CFB=<ABF. Let CH be the tangent  to circle AEC (H on AB)

<ECH=<EFC=<ABF. ---> BCEH are concyclic <CEB=<CHB=90