Cho tam giác ABC. Gọi H là trực tâm của tam giác và M là trung điểm của cạnh BC. Chứng minh rằng \(\overrightarrow{MH}.\overrightarrow{MA}=\dfrac{1}{4}BC^2\) ?
A B C H M Có \(\overrightarrow{MH}=-\overrightarrow{HM}=\dfrac{-1}{2}\left(\overrightarrow{HB}+\overrightarrow{HC}\right)\); \(\overrightarrow{MA}=-\overrightarrow{AM}=\dfrac{-1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\). Vì vậy: \(\overrightarrow{MH}.\overrightarrow{MA}=\dfrac{-1}{2}\left(\overrightarrow{HB}+\overrightarrow{HC}\right).\dfrac{-1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\) \(=\dfrac{1}{4}\left(\overrightarrow{HB}.\overrightarrow{AB}+\overrightarrow{HB}.\overrightarrow{AC}+\overrightarrow{HC}.\overrightarrow{AB}+\overrightarrow{HC}.\overrightarrow{AC}\right)\) \(=\dfrac{1}{4}\left(\overrightarrow{CH}.\overrightarrow{AC}+\overrightarrow{BH}.\overrightarrow{AB}\right)\) (Do H là trực tâm tam giác ABC). \(=\dfrac{1}{4}\left[\overrightarrow{CH}\left(\overrightarrow{AB}+\overrightarrow{BC}\right)+\overrightarrow{BH}\left(\overrightarrow{AB}+\overrightarrow{BC}\right)\right]\) \(=\dfrac{1}{4}\left(\overrightarrow{CH}.\overrightarrow{AB}+\overrightarrow{CH}.\overrightarrow{BC}+\overrightarrow{BH}.\overrightarrow{AB}+\overrightarrow{BH}.\overrightarrow{BC}\right)\) \(=\dfrac{1}{4}\left(\overrightarrow{CH}.\overrightarrow{BC}+\overrightarrow{BH}.\overrightarrow{BC}\right)\) ( do H là trực tâm tam giác ABC). \(=\dfrac{1}{4}\overrightarrow{BC}\left(\overrightarrow{BH}+\overrightarrow{HC}\right)\) \(=\dfrac{1}{4}\overrightarrow{BC}.\overrightarrow{BC}=\dfrac{1}{4}BC^2\).