Vector Algebra Question 208
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 = DT. The length AT is equal to
Options:
A) 2BC
B) 3 BCE
C) $ \frac{4}{3} $ BC
D) none of these
Show Answer
Answer:
Correct Answer: A
Solution:
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{a}|=|\overrightarrow{b}|=|\overrightarrow{c}|=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}=| -2\vec{a} |=2| {\vec{a}} |=2R. $ But $ BC=2R\sin A; $ therefore, AT=2BC
BETA
