a.
D chia CB theo tỉ số \(k=2\Rightarrow\)\(\overrightarrow{DC}=2\overrightarrow{DB}\)
\(\Rightarrow\overrightarrow{DC}-\overrightarrow{DB}=\overrightarrow{DB}\Rightarrow\overrightarrow{DC}+\overrightarrow{BD}=\overrightarrow{DB}\)
\(\Rightarrow\overrightarrow{BC}=\overrightarrow{DB}\Rightarrow\overrightarrow{BC}+\overrightarrow{BD}=\overrightarrow{0}\)
\(\Rightarrow\) B là trung điểm CD hay D là điểm đối xứng C qua B
Do M là trung điểm AB \(\Rightarrow\overrightarrow{MA}=-\overrightarrow{MB}\)
\(\overrightarrow{CA}.\overrightarrow{CB}=\left(\overrightarrow{CM}+\overrightarrow{MA}\right)\left(\overrightarrow{CM}+\overrightarrow{MB}\right)=\left(2\overrightarrow{CI}-\overrightarrow{MB}\right)\left(2\overrightarrow{CI}+\overrightarrow{MB}\right)\)
\(=4\overrightarrow{CI}^2-\overrightarrow{MB}^2=4CI^2-MB^2\)
b.
\(2\left(1-cos^2C\right)+3cosC=0\Leftrightarrow-2cos^2C+3cosC+2=0\Rightarrow\left[{}\begin{matrix}cosC=2>1\left(loại\right)\\cosC=-\dfrac{1}{2}\end{matrix}\right.\)
Mặt khác: \(cosC=\dfrac{AC^2+BC^2-AB^2}{2AC.BC}=\dfrac{20-AB^2}{16}=-\dfrac{1}{2}\)
\(\Rightarrow AB^2=28\Rightarrow AB=2\sqrt{7}\)
\(\Rightarrow cosB=\dfrac{AB^2+BC^2-AC^2}{2AB.BC}=\dfrac{5\sqrt{7}}{14}\)
\(\Rightarrow CH=BC.\sqrt{1-cos^2B}=\dfrac{2\sqrt{21}}{7}\)
\(BM=\dfrac{1}{2}AB=\sqrt{7}\Rightarrow CM=\sqrt{BM^2+BC^2-2BM.BC.cosB}=\sqrt{3}\)
Áp dụng công thức trung tuyến:
\(BI=\dfrac{\sqrt{2\left(BM^2+BC^2\right)-CM^2}}{2}=...\)
