如题所述
∵△ABM沿直线AM翻折后,点B恰好落在边AC的中点处,假设这个点是B′,作MN⊥AC,MD⊥AB,垂足分别为N,D.又∵Rt△ABC中,∠BAC=90°,AB=3,∴AB=AB′=3,DM=MN,AB′=B′C=3,S△BAC=S△BAM+S△MAC1/2×3×6=1/2×MD×3+1/2×6×MN,∴解得:MD=2,所以点M到AC的距离是2.