\(\overrightarrow{AB}+\overrightarrow{CD}=\overrightarrow{AB}+\overrightarrow{CB}+\overrightarrow{BD}=\overrightarrow{AB}+\overrightarrow{BD}+\overrightarrow{CB}=\overrightarrow{AD}+\overrightarrow{CB}\)

\(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}+\overrightarrow{OD}=\left(\overrightarrow{OE}+\overrightarrow{EA}\right)+\left(\overrightarrow{OF}+\overrightarrow{FB}\right)+\left(\overrightarrow{OE}+\overrightarrow{EC}\right)+\left(\overrightarrow{OF}+\overrightarrow{FD}\right)\)

\(=2\left(\overrightarrow{OE}+\overrightarrow{EF}\right)+\left(\overrightarrow{EA}+\overrightarrow{EC}\right)+\left(\overrightarrow{FB}+\overrightarrow{FD}\right)\)

\(=2.\overrightarrow{0}+\overrightarrow{0}+\overrightarrow{0}=\overrightarrow{0}\)

Tham khảo:

Dễ thấy: \(\overrightarrow {OA}  = \overrightarrow {OM}  + \overrightarrow {MA} \); \(\overrightarrow {OB}  = \overrightarrow {OM}  + \overrightarrow {MB} \)

Tương tự: \(\overrightarrow {OC}  = \overrightarrow {ON}  + \overrightarrow {NC} \); \(\overrightarrow {OD}  = \overrightarrow {ON}  + \overrightarrow {ND} \)

\(\begin{array}{l} \Rightarrow \overrightarrow {OA}  + \overrightarrow {OB}  + \overrightarrow {OC}  + \overrightarrow {OD}  = \left( {\overrightarrow {OM}  + \overrightarrow {MA} } \right) + \left( {\overrightarrow {OM}  + \overrightarrow {MB} } \right) + \left( {\overrightarrow {ON}  + \overrightarrow {NC} } \right) + \left( {\overrightarrow {ON}  + \overrightarrow {ND} } \right)\\ = \left( {\overrightarrow {OM}  + \overrightarrow {OM}  + \overrightarrow {MA}  + \overrightarrow {MB} } \right) + \left( {\overrightarrow {ON}  + \overrightarrow {ON}  + \overrightarrow {NC}  + \overrightarrow {ND} } \right)\\ = \overrightarrow {OM}  + \overrightarrow {OM}  + \overrightarrow {ON}  + \overrightarrow {ON} \\ = \left( {\overrightarrow {OM}  + \overrightarrow {ON} } \right) + \left( {\overrightarrow {OM}  + \overrightarrow {ON} } \right)\\ = \overrightarrow 0  + \overrightarrow 0 \\ = \overrightarrow 0 .\end{array}\)

\(\overrightarrow{OA'}+\overrightarrow{OB'}+\overrightarrow{OC'}\)

\(=\overrightarrow{OA}+\overrightarrow{AA'}+\overrightarrow{OB}+\overrightarrow{BB'}+\overrightarrow{OC}+\overrightarrow{CC'}\)

\(=\left(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\right)+\left(\overrightarrow{BA}+\overrightarrow{CB}+\overrightarrow{AC}\right)\)

\(=\left(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\right)+\overrightarrow{CC}=\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}\) (đpcm)

Ta có:

\(\overrightarrow {MN}  = \overrightarrow {MA}  + \overrightarrow {AD}  + \overrightarrow {DN} \)

Mặt khác: \(\overrightarrow {MN}  = \overrightarrow {MB}  + \overrightarrow {BC}  + \overrightarrow {CN} \)

\(\begin{array}{l} \Rightarrow 2\overrightarrow {MN}  = \overrightarrow {MA}  + \overrightarrow {AD}  + \overrightarrow {DN}  + \overrightarrow {MB}  + \overrightarrow {BC}  + \overrightarrow {CN} \\ \Leftrightarrow 2\overrightarrow {MN}  = \left( {\overrightarrow {MA}  + \overrightarrow {MB} } \right) + \left( {\overrightarrow {DN}  + \overrightarrow {CN} } \right) + \overrightarrow {BC}  + \overrightarrow {AD} \\ \Leftrightarrow 2\overrightarrow {MN}  = \overrightarrow 0  + \overrightarrow 0  + \overrightarrow {BC}  + \overrightarrow {AD} \\ \Leftrightarrow 2\overrightarrow {MN}  = \overrightarrow {BC}  + \overrightarrow {AD} \end{array}\)

Lại có: 

\(\overrightarrow {BC}  + \overrightarrow {AD}  = \overrightarrow {BD}  + \overrightarrow {DC}  + \overrightarrow {AD}  = \overrightarrow {AD}  + \overrightarrow {DC} + \overrightarrow {BD}  = \overrightarrow {AC}  + \overrightarrow {BD} .\)

Vậy \(\overrightarrow {BC}  + \overrightarrow {AD}  = 2\overrightarrow {MN}  = \;\overrightarrow {AC}  + \overrightarrow {BD} .\)

1.

Gọi G là trọng tâm tam giác

\(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{0}\)

\(\Leftrightarrow3\overrightarrow{OG}=\overrightarrow{0}\)

\(\Leftrightarrow O\equiv G\)

\(\Rightarrow O\) là trọng tâm tam giác ABC

\(\Rightarrow\Delta ABC\) đều

Gọi độ dài các cạnh tam giác là a

\(\overrightarrow{BN}.\overrightarrow{AM}=\dfrac{1}{4}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\left(\overrightarrow{BA}+\overrightarrow{BC}\right)=-\dfrac{1}{4}a^2-\dfrac{1}{8}a^2-\dfrac{1}{8}a^2+\dfrac{1}{2}a^2=0\)

Mặt khác \(\overrightarrow{BN}.\overrightarrow{AM}=BN.AM.cos\left(\overrightarrow{AM};\overrightarrow{BN}\right)\)

\(\Rightarrow BN.AM.cos\left(\overrightarrow{AM};\overrightarrow{BN}\right)=0\Rightarrow cos\left(\overrightarrow{AM};\overrightarrow{BN}\right)=0\Rightarrow\left(\overrightarrow{AM};\overrightarrow{BN}\right)=90^o\)

\(BD=\dfrac{AB}{cos45^o}=\dfrac{a}{\dfrac{\sqrt{2}}{2}}=a\sqrt{2}\)

\(\overrightarrow{BQ}.\overrightarrow{BP}=\dfrac{1}{4}\left(\overrightarrow{BA}+\overrightarrow{BD}\right)\left(\overrightarrow{BC}+\overrightarrow{BD}\right)\)

\(=\dfrac{1}{4}BA.BC.cos90^o+\dfrac{1}{4}BA.BD.cos45^o+\dfrac{1}{4}BD.BC.cos45^o+\dfrac{1}{4}BD^2\)

\(=\dfrac{1}{4}a^2+\dfrac{1}{4}a^2+\dfrac{1}{2}a^2=a^2\)

a) Gọi M là trung điểm của BC nên:

Ta có:

vì

Mặt khác, do D là trung điểm của đoạn AM nên

Khi đó:

b) Ta có:

luôn đúng theo câu a

Vậy: , với O là điểm tùy ý