<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
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