Lời giải

Ta cần chứng minh AM ⊥ DB Û \(\overrightarrow {AM} .\overrightarrow {BD} = 0\).

Vì M là trung điểm của HD nên \(2\overrightarrow {AM} = \overrightarrow {AH} + \overrightarrow {AD} \).

Ta có: \(\overrightarrow {BD} = \overrightarrow {BH} + \overrightarrow {HD} \).

Do đó \(2\overrightarrow {AM} .\overrightarrow {BD} = \left( {\overrightarrow {AH} + \overrightarrow {AD} } \right).\left( {\overrightarrow {BH} + \overrightarrow {HD} } \right)\)

                          \[ = \overrightarrow {AH} .\overrightarrow {BH} + \overrightarrow {AH} .\overrightarrow {HD} + \overrightarrow {AD} .\overrightarrow {BH} + \overrightarrow {AD} .\overrightarrow {HD} \]

                          \[ = \underbrace {\overrightarrow {AH} .\overrightarrow {BH} }_{ = 0\,\left( {do\,\,AH \bot BH} \right)} + \overrightarrow {AH} .\overrightarrow {HD} + \overrightarrow {AD} .\overrightarrow {BH} + \underbrace {\overrightarrow {AD} .\overrightarrow {HD} }_{ = 0\,\left( {do\,\,AD \bot HD} \right)}\]

                          \[ = \overrightarrow {AH} .\overrightarrow {HD} + \overrightarrow {AD} .\overrightarrow {BH} \]

                          \[ = \overrightarrow {AH} .\overrightarrow {HD} + \left( {\overrightarrow {AH} + \overrightarrow {HD} } \right).\overrightarrow {BH} \]

                          \[ = \overrightarrow {AH} .\overrightarrow {HD} + \overrightarrow {AH} .\overrightarrow {BH} + \overrightarrow {HD} .\overrightarrow {BH} \]

                          \[ = \overrightarrow {AH} .\overrightarrow {HD} + \underbrace {\overrightarrow {AH} .\overrightarrow {BH} }_{ = 0\,\left( {do\,\,AH \bot BH} \right)} + \overrightarrow {HD} .\overrightarrow {BH} \]

                          \[ = \overrightarrow {HD} \left( {\overrightarrow {AH} + \overrightarrow {BH} } \right)\]

                          \[ = \overrightarrow {HD} .\left( {\overrightarrow {AH} + \overrightarrow {HC} } \right)\,\,\,\left( {do\,\,\,\overrightarrow {BH} = \overrightarrow {HC} } \right)\]

                          \[ = \overrightarrow {HD} .\overrightarrow {AC} = 0\,\,\,\left( {doHD \bot AC} \right)\]

Do đó \(2\overrightarrow {AM} .\overrightarrow {BD} = 0 \Leftrightarrow \overrightarrow {AM} .\overrightarrow {BD} = 0\).

Vậy AM ⊥ DB.