<Tri-2025-2226>, hint

Let M be the midpoint of BC.

Since circumcenter of ABC and incenter of DBC are identical, O=I. OM=IM ㅗBC.

Triangles IMB and IMC are congruent. Hence <DBC(2 <IBM)=<DCB(2<ICM), 

Finally <ABC=<ACD and triangles ABC and ACD are similar ,and so AC/AB=AD/AC