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A B C H D M
Tam giác ABC cân tại A, H là trung điểm của BC nên \(AH\perp BC\).
Có \(\overrightarrow{AM}.\overrightarrow{BD}=\dfrac{1}{2}\left(\overrightarrow{AH}+\overrightarrow{AD}\right)\left(\overrightarrow{BH}+\overrightarrow{HD}\right)\)
\(=\dfrac{1}{2}\left(\overrightarrow{AH}.\overrightarrow{BH}+\overrightarrow{AH}.\overrightarrow{HD}+\overrightarrow{AD}.\overrightarrow{BH}+\overrightarrow{AD}.\overrightarrow{HD}\right)\)
\(=\dfrac{1}{2}\left(\overrightarrow{AH}.\overrightarrow{HD}+\overrightarrow{AD}.\overrightarrow{BH}\right)\) (do \(AH\perp BC\) )
\(=\dfrac{1}{2}\overrightarrow{AH}.\left(\overrightarrow{BH}+\overrightarrow{HD}\right)+\dfrac{1}{2}\left(\overrightarrow{AH}+\overrightarrow{HD}\right).\overrightarrow{BH}\)
\(=\dfrac{1}{2}\overrightarrow{AH}.\overrightarrow{BH}+\dfrac{1}{2}\overrightarrow{AH}.\overrightarrow{HD}+\dfrac{1}{2}\overrightarrow{AH}.\overrightarrow{BH}+\dfrac{1}{2}\overrightarrow{HD}.\overrightarrow{BH}\)
\(=\dfrac{1}{2}\overrightarrow{AH}.\overrightarrow{HD}+\dfrac{1}{2}\overrightarrow{HD}.\overrightarrow{BH}\) ( do \(AH\perp BC\) )
\(=\dfrac{1}{2}\overrightarrow{HD}\left(\overrightarrow{AH}+\overrightarrow{BH}\right)\)
\(=\dfrac{1}{2}\overrightarrow{HD}\left(\overrightarrow{AH}+\overrightarrow{HC}\right)\) ( doM là trung điểm của BC).
\(=\dfrac{1}{2}\overrightarrow{HD}.\overrightarrow{AC}\)
\(=0\) (Do \(HD\perp AC\) )
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 \(MP\perp BC\).