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16 tháng 10 2019

AB+CD=AD+CB

<=> AB-AD=CB-CD

<=> DB=DB

a: \(\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DE}\)

\(=\overrightarrow{AC}+\overrightarrow{CD}+\overrightarrow{DE}\)

\(=\overrightarrow{AC}+\overrightarrow{CE}=\overrightarrow{AE}\)

b: \(\overrightarrow{AC}+\overrightarrow{BE}+\overrightarrow{CB}+\overrightarrow{DA}\)

\(=\overrightarrow{AC}+\overrightarrow{CB}+\overrightarrow{BE}+\overrightarrow{DA}=\overrightarrow{AB}+\overrightarrow{BE}+\overrightarrow{DA}\)

\(=\overrightarrow{DA}+\overrightarrow{AE}=\overrightarrow{DE}\)

c: \(\overrightarrow{AB}+\overrightarrow{DC}+\overrightarrow{BD}+\overrightarrow{CA}\)

\(=\overrightarrow{AB}+\overrightarrow{BD}+\overrightarrow{DC}+\overrightarrow{CA}\)

\(=\overrightarrow{AD}+\overrightarrow{DA}=\overrightarrow{0}\)

d: \(\overrightarrow{BC}+\overrightarrow{AD}+\overrightarrow{CD}+\overrightarrow{DA}\)

\(=\overrightarrow{AD}+\overrightarrow{BC}+\overrightarrow{CA}=\overrightarrow{AD}+\overrightarrow{BA}=\overrightarrow{BD}\)

22 tháng 7 2018

a) ta có : \(\overrightarrow{AB}-\overrightarrow{CD}=\overrightarrow{AB}+\overrightarrow{DC}=\overrightarrow{AP}+\overrightarrow{PQ}+\overrightarrow{QB}+\overrightarrow{DP}+\overrightarrow{PQ}+\overrightarrow{QC}\)

\(=2\overrightarrow{PQ}+\left(\overrightarrow{AP}+\overrightarrow{DP}\right)+\left(\overrightarrow{QB}+\overrightarrow{QC}\right)=2\overrightarrow{PQ}\) ..................(1)

\(\overrightarrow{AC}-\overrightarrow{BD}=\overrightarrow{AC}+\overrightarrow{DB}=\overrightarrow{AP}+\overrightarrow{PQ}+\overrightarrow{QC}+\overrightarrow{DP}+\overrightarrow{PQ}+\overrightarrow{QB}\)

\(=2\overrightarrow{PQ}+\left(\overrightarrow{AP}+\overrightarrow{DP}\right)+\left(\overrightarrow{QB}+\overrightarrow{QC}\right)=2\overrightarrow{PQ}\) ..................(2)

từ (1) (2) ta có : \(\overrightarrow{AB}-\overrightarrow{CD}=\overrightarrow{AC}-\overrightarrow{BD}=2\overrightarrow{PQ}\left(đpcm\right)\)