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Ta có: IJ−→=IA−→+AB−→−+BJ−→IJ→=IA→+AB→+BJ→
IJ−→=ID−→+DC−→−+CJ−→IJ→=ID→+DC→+CJ→
⇒IJ−→=12(AB−→−+DC−→−)⇒IJ→=12(AB→+DC→)
Xét:
HK−→−.IJ→=12(OK−→−−OH−→−).(AB−→−+DC−→−)=12(OK−→−.AB−→−+OK−→−.DC−→−−OH−→−.AB−→−−OH−→−.DC−→−)=12(OK−→−.AB−→−−OH−→−.DC−→−)=12[(OC−→−+CK−→−).(OB−→−−OA−→−)−(OA−→−+AH−→−).(OC−→−−OD−→−)]=12[(OB−→−−OA−→−−AH−→−).OC−→−−(CK−→−+OC−→−−OD−→−).OA−→−]=12[(HA−→−+AO−→−+OB−→−).OC−→−−(DO−→−+OC−→−+CK−→−).OA−→−]=12(HB−→−.OC−→−−DK−→−.OA−→−)=0⇔HK⊥IJ
A B C D P M
a) \(\overrightarrow{MP}.\overrightarrow{BC}=\dfrac{1}{2}\left(\overrightarrow{MA}+\overrightarrow{MD}\right).\left(\overrightarrow{BM}+\overrightarrow{MC}\right)\)
\(=\dfrac{1}{2}\left(\overrightarrow{MA}.\overrightarrow{BM}+\overrightarrow{MA}.\overrightarrow{MC}+\overrightarrow{MD}.\overrightarrow{BM}+\overrightarrow{MD}.\overrightarrow{MC}\right)\)
\(=\dfrac{1}{2}\left(\overrightarrow{MA}.\overrightarrow{BM}+\overrightarrow{MA}.\overrightarrow{MC}-\overrightarrow{MB}.\overrightarrow{MD}+\overrightarrow{MD}.\overrightarrow{MC}\right)\)
\(=\dfrac{1}{2}\left(\overrightarrow{MA}.\overrightarrow{BM}+\overrightarrow{MD}.\overrightarrow{MC}\right)\)
\(=\dfrac{1}{2}\left(0+0\right)=0\) (vì \(AC\perp BD\) nên \(\overrightarrow{MA}.\overrightarrow{BM}=0;\overrightarrow{MD}.\overrightarrow{MC}=0\)).
Vậy \(\overrightarrow{MP}.\overrightarrow{BC}=0\) nên \(MP\perp BC\).