# Bài 23 trang 24 Sách giáo khoa (SGK) Hình học 10 Nâng cao

##### Hướng dẫn giải

Theo quy tắc ba điểm, ta có

\eqalign{ & \overrightarrow {AC} + \overrightarrow {BD} = \left( {\overrightarrow {AM} + \overrightarrow {MN} + \overrightarrow {NC} } \right) + \left( {\overrightarrow {BM} + \overrightarrow {MN} + \overrightarrow {ND} } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\overrightarrow {MN} + \left( {\overrightarrow {AM} + \overrightarrow {BM} } \right) + \left( {\overrightarrow {NC} + \overrightarrow {ND} } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\overrightarrow {MN} + \overrightarrow 0 + \overrightarrow 0 = 2\overrightarrow {MN} \cr & \overrightarrow {AD} + \overrightarrow {BC} = \left( {\overrightarrow {AM} + \overrightarrow {MN} + \overrightarrow {ND} } \right) + \left( {\overrightarrow {BM} + \overrightarrow {MN} + \overrightarrow {NC} } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\overrightarrow {MN} + \left( {\overrightarrow {AM} + \overrightarrow {BM} } \right) + \left( {\overrightarrow {NC} + \overrightarrow {ND} } \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 2\overrightarrow {MN} + \overrightarrow 0 + \overrightarrow 0 = 2\overrightarrow {MN} \cr}

Vậy $$2\overrightarrow {MN} = \overrightarrow {AC} + \overrightarrow {BD} = \overrightarrow {AD} + \overrightarrow {BC} .$$