<Tri-2023-2052>, hint

Consider Menelaus theorem to triangle XYM and line AM .

YM/MD x DA/AX x XN/NY=1 OY/OA xDA /AX== x1/1=1

OY/YA=R cosA/ cos (C-B) R.Since triangles AEX and ABY are similar

AX/AE=AY/AB, AX=c cosA x AD/ cos(C-B) c=cosA AD/ cos (C-B)

Hence AS/AX=cos(C-B)/cosA

Hence YM/MD xDA/AX c XN/NY=cosA/cos(C-B) x cos(C-B)/ cosA x 1/1=1.