<Cir-2025-791>, hint

Let <AOB=2z, <OAK=y, <OBK=x A circle inscribed in angle O touches A, B

Hence <OAK=<ABK=y <OBK=<BAK=x. Since LK//OA <KLB=<KMB=<AOB=2z, L,K,B,M are cyclic

<KLB=<KMB=2z=<AOB  Thus A,O, M, B are cyclic.

OB cuts MK at N, <OMA=<OBA=x+y, <BAM=x=<OBK=<OMK,

<OAM=<LKM=y, <OML=<OMN-<LMN=(x+y-x)=y=<MKL. Hence OM tangent to  circle w.

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