<Tri-2024-2095>, hint

Suppose AB<AC and line through M parallel to YZ meets AC in N and AB  in N',

(AN'>AB). Angle bisector of A cuts BB' in D,  BB' is parallel to YZ, B' on AC.

Triangles ABD and ANM are congruent . DM//AC, DB'=MN=BD, <ABD=<ANM.

Let K be he intersection of lines ML and BA, then  <NLM=<ALK=A/2. Hence <AKA=A/2

KA=AL=DM=NB'=NC (NN'//BB'. BM=MC, AD=//LM)