<Cir-2026-796>, hint

Let XTY be tangent to both circles

<ATX=W=<ABT, <PTX=Z=<PCT (TB cuts small circle at C), <ATP=V, Z=W+V

<PCT=<PTX=W+V, Hence <BPC=<PCT-<PBC=W+V-W=V, which is <PTC( ATB is tangent  to

circle TPC). Hence <ATP=<PTC=V.

<Cir-2026- 796>, general, bisector>

<Singapore Senior Math Olympiad 2006>

<Qua-2026-804>, hint

 

<Qua-2026-804 >, general, cyclic

<Singapore Senior Math Olympiad 2006>

<Qua-202-803>, hint

Consider a triangle ABD.

Let D' on BD such that JD'=//ID , then triangles JBD' and CEF are  congruent , 

<JBD'=<CEF  .Hence BD//EF. Hence KL is parallel to EF and BD.  ID=FC