A) 2BC
B) 3BC
C) \[\frac{4}{3}\]BC
D) none of these
Correct Answer: A
Solution :
[a] Let the origin of reference be the circumcenter of the triangle. Let \[\overrightarrow{OA}=\overrightarrow{a}=\overrightarrow{OB}=\overrightarrow{b},\overrightarrow{OC}=\overrightarrow{c}\]and \[\overrightarrow{OT}=\overrightarrow{t}\] Then \[\overrightarrow{\left| a \right|}=\overrightarrow{\left| b \right|}=\overrightarrow{\left| c \right|}=R\](circumradius) Again \[\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}=\overrightarrow{OA}+2\overrightarrow{OD}=\overrightarrow{OA}+\overrightarrow{AH}=\overrightarrow{OH}\] Therefore, the P.V. of H is \[\vec{a}+\vec{b}+\vec{c}\]. Since D is the midpoint of HT, we have \[\frac{\vec{a}+\vec{b}+\vec{c}+\vec{t}}{2}=\frac{\vec{b}+\vec{c}}{2}\Rightarrow \vec{t}=-\vec{a}\] \[\therefore \overrightarrow{AT}=-2\overrightarrow{a}\] or \[\overrightarrow{AT}=\left| -2\vec{a} \right|=2\left| {\vec{a}} \right|=2R.\] But \[BC=2R\sin A=R;\] therefore, AT=2BCYou need to login to perform this action.
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