question_answer

In triangle ABC,\[\angle A=30{}^\circ \], H is the orthocentre and D is the midpoint of BC. Segment HD is produced to T such that HD = DF. The length AT is equal to

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=2BC