a) \(\overrightarrow {AC}  + \overrightarrow {BD} = \overrightarrow {AM}  + \overrightarrow {MN}  + \overrightarrow {NC}  + \overrightarrow {BM}  + \overrightarrow {MN}  + \overrightarrow {ND}  \\=  \left( {\overrightarrow {AM}  + \overrightarrow {BM} } \right) + \left( {\overrightarrow {MN}  + \overrightarrow {MN} } \right) + \left( {\overrightarrow {NC}  + \overrightarrow {ND} } \right) \\=  \overrightarrow 0  + 2\overrightarrow {MN}  + \overrightarrow 0  = 2\overrightarrow {MN} \) (đpcm)                                                             

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

\(\)\(\overrightarrow {BC}  + \overrightarrow {AD}  = \overrightarrow {BM}  + \overrightarrow {MN}  + \overrightarrow {NC}  + \overrightarrow {AM}  + \overrightarrow {MN}  + \overrightarrow {ND} \)

\(\left( {\overrightarrow {BM}  + \overrightarrow {AM} } \right) + \left( {\overrightarrow {MN}  + \overrightarrow {MN} } \right) + \left( {\overrightarrow {NC}  + \overrightarrow {ND} } \right) = 2\overrightarrow {MN} \)

Mặt khác ta có: \(\overrightarrow {AC}  + \overrightarrow {BD}  = 2\overrightarrow {MN} \)

Suy ra \(\overrightarrow {AC}  + \overrightarrow {BD}  = \overrightarrow {BC}  + \overrightarrow {AD} \)

Cách 2: 

\(\begin{array}{l}
\overrightarrow {AC} + \overrightarrow {BD} = \overrightarrow {BC} + \overrightarrow {AD} \\
\Leftrightarrow \overrightarrow {AC} - \overrightarrow {AD} = \overrightarrow {BC} - \overrightarrow {BD} \\
\Leftrightarrow \overrightarrow {DC} = \overrightarrow {DC} (đpcm)
\end{array}\)

Tham khảo:

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

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

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

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

a)
MN là đường trung bình của tam giác ABC nên \(\overrightarrow{MN}=\dfrac{1}{2}\overrightarrow{AC}\).
QP là đường trung bình của tam giác ABC nên \(\overrightarrow{QP}=\dfrac{1}{2}\overrightarrow{AC}\).
Vậy \(\overrightarrow{MN}=\overrightarrow{QP}\).
b) Giả sử:
\(\overrightarrow{MP}=\overrightarrow{MN}+\overrightarrow{MQ}\Leftrightarrow\overrightarrow{MP}-\overrightarrow{MN}-\overrightarrow{MQ}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{MP}+\overrightarrow{NM}+\overrightarrow{QM}=\overrightarrow{0}\)
\(\Leftrightarrow\left(\overrightarrow{QM}+\overrightarrow{MP}\right)+\overrightarrow{NM}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{QP}+\overrightarrow{NM}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{QP}-\overrightarrow{MN}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{QP}-\overrightarrow{QP}=\overrightarrow{0}\)
\(\Leftrightarrow\overrightarrow{0}=\overrightarrow{0}\) ( Điều giả sử đúng).
Vậy \(\overrightarrow{MP}=\overrightarrow{MN}+\overrightarrow{MQ}.\)

a) Chữa đề: \(\overrightarrow{CA}+\overrightarrow{DB}=\overrightarrow{CB}+\overrightarrow{DA}=2\overrightarrow{NM}\)

\(Ta\text{ }có:\overrightarrow{CA}+\overrightarrow{DB}=\overrightarrow{CB}+\overrightarrow{BA}+\overrightarrow{DA}+\overrightarrow{AB}\\ =\overrightarrow{CB}+\overrightarrow{DA}+\left(\overrightarrow{BA}+\overrightarrow{AB}\right)=\overrightarrow{CB}+\overrightarrow{DA}\)

\(\)\(\overrightarrow{CA}+\overrightarrow{DB}=\overrightarrow{CA}+\overrightarrow{CB}+\overrightarrow{DC}\\ =2\overrightarrow{CM}+2\overrightarrow{NC}=2\left(\overrightarrow{NC}+\overrightarrow{CM}\right)=2\overrightarrow{NM}\)

Vậy \(\overrightarrow{CA}+\overrightarrow{DB}=\overrightarrow{CB}+\overrightarrow{DA}=2\overrightarrow{NM}\)

\(\text{b) }\overrightarrow{AD}+\overrightarrow{BD}+\overrightarrow{AC}+\overrightarrow{BC}=-\left(\overrightarrow{DA}+\overrightarrow{DB}+\overrightarrow{CA}+\overrightarrow{CB}\right)\\ =-\left[\left(\overrightarrow{DA}+\overrightarrow{DB}\right)+\left(\overrightarrow{CA}+\overrightarrow{CB}\right)\right]\\ =-\left(2\overrightarrow{DM}+2\overrightarrow{CM}\right)=2\left(\overrightarrow{MD}+\overrightarrow{MC}\right)=4\left(\overrightarrow{MN}\right)\)

\(\text{c) }2\left(\overrightarrow{AB}+\overrightarrow{AI}+\overrightarrow{NA}+\overrightarrow{DA}\right)\\ =2\left[\left(\overrightarrow{AB}+\overrightarrow{DA}\right)+\left(\overrightarrow{AI}+\overrightarrow{NA}\right)\right]\\ =2\left[\left(\overrightarrow{AB}+\overrightarrow{BA}+\overrightarrow{DB}\right)+\overrightarrow{NI}\right]=2\left(\overrightarrow{DB}+\overrightarrow{NI}\right)\)

Mà IN là dường trung bình \(\Delta BCD\)

\(\Rightarrow\left\{{}\begin{matrix}IN//BD\\IN=\frac{1}{2}BD\end{matrix}\right.\Rightarrow\overrightarrow{IN}=\frac{1}{2}\overrightarrow{BD}\\ \Rightarrow2\left(\overrightarrow{AB}+\overrightarrow{AI}+\overrightarrow{NA}+\overrightarrow{DA}\right)\\ =2\left(\overrightarrow{DB}+\overrightarrow{NI}\right)=2\left(\overrightarrow{DB}+\frac{1}{2}\overrightarrow{DB}\right)=2\cdot\frac{3}{2}\overrightarrow{DB}=3\overrightarrow{DB}\)