Cho hình thoi \(ABCD\) có cạnh bằng 2 và góc \(B\) bằng ${60}^{°}$. Khi đó:

a) $(\overrightarrow{AB},\overrightarrow{AC})={60}^{°}$

b) $(\overrightarrow{AB},\overrightarrow{DA})={30}^{°}$

c) \(\overrightarrow {DA}  \cdot \overrightarrow {DC}  = 3\)

d) \(\overrightarrow {OB}  \cdot \overrightarrow {BA}  =  - 3\)

Ta có: \(\overrightarrow {AE}  = \overrightarrow {AB}  + \overrightarrow {BE}  = \overrightarrow {AB}  + \frac{1}{2}\overrightarrow {BC}  = \overrightarrow {AB}  + \frac{1}{2}\overrightarrow {AD} \).

\(\begin{array}{l}\overrightarrow {AF}  = \overrightarrow {AB}  + \overrightarrow {BF}  = \overrightarrow {AB}  + \frac{3}{4}\overrightarrow {BD}  = \overrightarrow {AB}  + \frac{3}{4}(\overrightarrow {AD}  - \overrightarrow {AB} ) = \frac{1}{4}\overrightarrow {AB}  + \frac{3}{4}\overrightarrow {AD} .\\\overrightarrow {EF}  = \overrightarrow {AF}  - \overrightarrow {AE}  = \left( {\frac{1}{4}\overrightarrow {AB}  + \frac{3}{4}\overrightarrow {AD} } \right) - \left( {\overrightarrow {AB}  + \frac{1}{2}\overrightarrow {AD} } \right) = \frac{{ - 3}}{4}\overrightarrow {AB}  + \frac{1}{4}\overrightarrow {AD} .\end{array}\)

Ta có: \(\overrightarrow {AF}  \cdot \overrightarrow {EF}  = \left( {\frac{1}{4}\overrightarrow {AB}  + \frac{3}{4}\overrightarrow {AD} } \right) \cdot \left( {\frac{{ - 3}}{4}\overrightarrow {AB}  + \frac{1}{4}\overrightarrow {AD} } \right)\)

\( = \frac{{ - 3}}{{16}}{\overrightarrow {AB} ^2} - \frac{1}{2}\overrightarrow {AB}  \cdot \overrightarrow {AD}  + \frac{3}{{16}}{\overrightarrow {AD} ^2} = 0 \Rightarrow AF \bot EF{\rm{. }}\)

Ta có: \({\overrightarrow {AF} ^2} = {\left( {\frac{1}{4}\overrightarrow {AB}  + \frac{3}{4}\overrightarrow {AD} } \right)^2} = \frac{1}{{16}}{\overrightarrow {AB} ^2} + \frac{3}{8}\overrightarrow {AB}  \cdot \overrightarrow {AD}  + \frac{9}{{16}}{\overrightarrow {AD} ^2} = \frac{5}{8}{\overrightarrow {AB} ^2}\).

\({\overrightarrow {EF} ^2} = {\left( {\frac{{ - 3}}{4}\overrightarrow {AB}  + \frac{1}{4}\overrightarrow {AD} } \right)^2} = \frac{9}{{16}}{\overrightarrow {AB} ^2} - \frac{3}{8}\overrightarrow {AB}  \cdot \overrightarrow {AD}  + \frac{1}{{16}}{\overrightarrow {AD} ^2} = \frac{5}{8}{\overrightarrow {AB} ^2}.\)

\( \Rightarrow A{F^2} = E{F^2} = \frac{5}{8}{\overrightarrow {AB} ^2} \Rightarrow AF = EF\). Vậy tam giác \(AEF\) vuông cân tại \(F\).

Chú ý: Ta có thể chứng minh tam giác \(AEF\) vuông bằng định lí Pythagore.

Ta cần chứng minh: \(\overrightarrow {AM}  \cdot \overrightarrow {BD}  = 0\). Ta có: \(\overrightarrow {BD}  = \overrightarrow {BH}  + \overrightarrow {HD}  = \overrightarrow {HC}  + \overrightarrow {HD} ;\overrightarrow {AM}  = \frac{1}{2}(\overrightarrow {AH}  + \overrightarrow {AD} )\)

Do đó: \(\overrightarrow {AM}  \cdot \overrightarrow {BD}  = \frac{1}{2}(\overrightarrow {AH}  + \overrightarrow {AD} )(\overrightarrow {HC}  + \overrightarrow {HD} )\)\( = \frac{1}{2}(\overrightarrow {AH}  \cdot \overrightarrow {HC}  + \overrightarrow {AH}  \cdot \overrightarrow {HD}  + \overrightarrow {AD}  \cdot \overrightarrow {HC}  + \overrightarrow {AD}  \cdot \overrightarrow {HD} )\),

mà \(\left\{ {\begin{array}{*{20}{l}}{\overrightarrow {AH}  \cdot \overrightarrow {HC}  = 0({\rm{ do }}AH \bot BC)}\\{\overrightarrow {AD}  \cdot \overrightarrow {HD}  = 0({\rm{ do }}HD \bot AC)}\end{array}} \right.\)\( \Rightarrow \overrightarrow {AM}  \cdot \overrightarrow {BD}  = \frac{1}{2}(\overrightarrow {AH}  \cdot \overrightarrow {HD}  + \overrightarrow {AD}  \cdot \overrightarrow {HC} )\)

\( = \frac{1}{2}[\overrightarrow {AH}  \cdot \overrightarrow {HD}  + (\overrightarrow {AH}  + \overrightarrow {HD} ) \cdot \overrightarrow {HC} ]\)

\( = \frac{1}{2}(\overrightarrow {AH}  \cdot \overrightarrow {HD}  + \underbrace {\overrightarrow {AH}  \cdot \overrightarrow {HC} }_0 + \overrightarrow {HD}  \cdot \overrightarrow {HC} ) = \frac{1}{2}\overrightarrow {HD}  \cdot (\overrightarrow {AH}  + \overrightarrow {HC} ) = \frac{1}{2}\overrightarrow {HD}  \cdot \overrightarrow {AC}  = 0\).

Vậy \(AM \bot DB\).