Chứng minh rằng tứ giác BMNK là hình bình hành

$$\begin{gathered} \mathrm{Cho} ∆ \mathrm{nhọn} \mathrm{ABC}, \mathrm{AB}<\mathrm{AC}. \mathrm{Gọi} \mathrm{M},\mathrm{N},\mathrm{K} \mathrm{lần} \mathrm{lượt} \mathrm{là} \mathrm{trung} \mathrm{điểm} \mathrm{của} \mathrm{AB}, \mathrm{AC}, \mathrm{BC}. \mathrm{Kẻ} \mathrm{AH}\perp \mathrm{BC}, \mathrm{H}\in \mathrm{BC} \\ \mathrm{a}/ \mathrm{CMR}: \mathrm{Tứ} \mathrm{giác} \mathrm{BMNK} \mathrm{là} \mathrm{hbh} \\ \mathrm{b}/ \mathrm{CMR}: \mathrm{Tứ} \mathrm{giác} \mathrm{MNKH} \mathrm{là} \mathrm{ht} \mathrm{cân} \end{gathered}$$