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Ta có: \(AC^2+BD^2=\left(\overrightarrow{AB}+\overrightarrow{AD}\right)^2+\left(\overrightarrow{BC}+\overrightarrow{BA}\right)^2\)
\(=AB^2+AD^2+2\overrightarrow{AB}.\overrightarrow{AD}+BC^2+BA^2+2\overrightarrow{BA}.\overrightarrow{BC}\)
\(=AB^2+AD^2+BC^2+AD^2+2\overrightarrow{AB}\left(\overrightarrow{AD}-\overrightarrow{BC}\right)\)
\(=AB^2+AD^2+BC^2+AD^2\)
\(AB^2+CD^2-\left(BC^2+DA^2\right)=\overrightarrow{AB}^2+\overrightarrow{CD}^2-\overrightarrow{BC}^2-\overrightarrow{AD}^2\)
\(=\left(\overrightarrow{AB}+\overrightarrow{AD}\right)\left(\overrightarrow{AB}-\overrightarrow{AD}\right)+\left(\overrightarrow{CD}-\overrightarrow{BC}\right)\left(\overrightarrow{CD}+\overrightarrow{BC}\right)\)
\(=\overrightarrow{DB}\left(\overrightarrow{AB}+\overrightarrow{AD}\right)+\overrightarrow{DB}\left(\overrightarrow{BC}+\overrightarrow{DC}\right)\)
\(=\overrightarrow{DB}\left(\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{BC}+\overrightarrow{DC}\right)\)
\(=2\overrightarrow{AC}.\overrightarrow{DB}\) (đpcm)
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:
\(\begin{array}{l}
\overrightarrow {AC} + \overrightarrow {BD} = \overrightarrow {BC} + \overrightarrow {AD} \\
\Leftrightarrow \overrightarrow {AC} - \overrightarrow {AD} = \overrightarrow {BC} - \overrightarrow {BD} \\
\Leftrightarrow \overrightarrow {DC} = \overrightarrow {DC} (đpcm)
\end{array}\)
M là trung điểm AB \(\Rightarrow\overrightarrow{IM}=\dfrac{1}{2}\left(\overrightarrow{IA}+\overrightarrow{IB}\right)\)
\(\Rightarrow2\overrightarrow{IM}.\overrightarrow{DC}=\left(\overrightarrow{IA}+\overrightarrow{IB}\right).\left(\overrightarrow{DI}+\overrightarrow{IC}\right)=\overrightarrow{IA}.\overrightarrow{DI}+\overrightarrow{IB}.\overrightarrow{IC}+\overrightarrow{IA}.\overrightarrow{IC}+\overrightarrow{IB}.\overrightarrow{DI}\)
\(=\overrightarrow{IA}.\overrightarrow{IC}+\overrightarrow{IB}.\overrightarrow{DI}=-IA.IC+IB.DI\)
Mặt khác do 2 tam giác vuông DIC và AIB đồng dạng (\(\widehat{IAB}=\widehat{IDC}\) cùng chắn BC)
\(\Rightarrow\dfrac{IA}{ID}=\dfrac{IB}{IC}\Rightarrow IA.IC=IB.ID\Rightarrow-IA.IC+IB.ID=0\)
\(\Rightarrow2\overrightarrow{IM}.\overrightarrow{DC}=0\Rightarrow IM\perp DC\)
