<Qua-2026-813>, hint

Let <BAD=2x, then <BAK=<BKA=<CKL=<CLK=x 

Let O be the circumcenter of triangle CKL, OK=OC (radius), <KOC=2KLC=2x.

BK=BA=CD, <OCD=<BCD+<OCK=2x+90-x=90+x, <OKB=180-<OKC(90-x)=90+x

Hence triangles OCD and OKB are congruent and <OBC=<ODC and O,B,C,D are cyclic.