Đặt \(AB = AC = AD=a\)

a) \(\overrightarrow{AB}.\overrightarrow{CD}=\overrightarrow{AB}.(\overrightarrow{AD}-\overrightarrow{AC})\)

\(=\overrightarrow{AB}.\overrightarrow{AD}-\overrightarrow{AB}.\overrightarrow{AC}\)

\(=AB.AD.\cos\widehat{BAD}-AB.AC.\cos\widehat{BAC} =a.a.cos60^0-a.a..cos60^0=0\)

 Cộng (1) với (2) theo vế với vế ta được:

$2.\overrightarrow{MN}=(\overrightarrow{MA}+\overrightarrow{MB})+(\overrightarrow{AD}+\overrightarrow{BC})+(\overrightarrow{DN}+\overrightarrow{CN})$

=> \(\overrightarrow{MN}=\frac{1}{2}(\overrightarrow{AD}+\overrightarrow{BC}\) (do M là trung điểm AB nên $\overrightarrow{MA}+\overrightarrow{MB}=\overrightarrow{0}$, N là trung điểm CD nên $\overrightarrow{DN}+\overrightarrow{CN}=\overrightarrow{0}$)

=> \(\overrightarrow{MN}=\frac{1}{2}(\overrightarrow{AD}+\overrightarrow{AC}-\overrightarrow{AB}).\)

=> \(\overrightarrow{AB}.\overrightarrow{MN}=\frac{1}{2}.\overrightarrow {AB} .(\overrightarrow {AD}  + \overrightarrow {AC}  - \overrightarrow {AB} )\)

\(= {1 \over 2}(\overrightarrow{AB}.\overrightarrow{AD}+\overrightarrow{AB}.\overrightarrow{AC}-\overrightarrow{AB}^{2})\)

\(= {1 \over 2}(AB.AD.\cos\widehat{BAD}+AB.AC.\cos\widehat{BAC}-AB^2)\)

\(={1 \over 2}(a.a.\cos60^0+a.a.\cos60^0-a^2)\)

\(={1 \over 2}\left({1 \over 2}a^2+{1 \over 2}a^2-a^2\right)=0\) \(\Rightarrow  AB ⊥ MN\).

Chứng minh tương tự ta được: \(CD ⊥ MN\).