<Qua-2024-742> hint

AD//BC ---><DAC=<BCA=p=<ACD (AC is angle bisector of BCD)

--->AD=DC=AO. Let <ADO=q, <ADO=<AOD=<COB=<CBO=q

Hence q=2p In triangle AOD  2q+p=180, p=36, q=72

Since DB=DC=DA, Let <OAB=s, then 2(p+s)+q---> s=18

Hence <ABC=s+p+q=18+36+72=126.