Kẻ HP//AC(P∊BM)
=> \(\dfrac{AM}{HP}=\dfrac{DA}{DH}=1\)(do D là trung điểm AH) ; 

     \(\dfrac{HP}{MC}=\dfrac{BH}{BC}\)

=>\(\dfrac{AM}{MC}=\dfrac{AM}{HP}.\dfrac{HP}{MC}=\dfrac{BH}{BC}\)

cosB=\(\dfrac{BH}{BA}=\dfrac{BA}{BC}\)

=>\(cos^2B=\dfrac{BH}{BA}.\dfrac{BA}{BC}=\dfrac{BH}{BC} \)

\(=>cos^2B=\dfrac{AM}{MC}\)