Question

In triangle $ABC$, $\angle A={30}^o$, $H$ 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

• A. $2BC$
• B. $3BC$
• C. $\frac{4}{3}BC$
• D. None of these

Answer 1

The correct option is A $2BC$

Let us assume that the circumcenter of $△ABC$ is $O$ and is situated at the origin of the coordinate plane. Hence, the vector notation for all the vertices is

$\overrightarrow{OA}=\overrightarrow{a}$

$\overrightarrow{OB}=\overrightarrow{b}$

$\overrightarrow{OC}=\overrightarrow{c}$

Also, as $O$ is the circumcenter of $△ABC$,

$\overrightarrow{a}=\overrightarrow{b}=\overrightarrow{c}=R$ (circum-radius)

Since $D$ is mid-point of $BC$, $\overrightarrow{d}=\frac{\overrightarrow{b}+\overrightarrow{c}}{2}$

$\therefore \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{a}+2\overrightarrow{d}$

Also, $2\overrightarrow{OD}=\overrightarrow{AH}$ as $H$ is the orthocentre and it divides height in 2:1 ratio.

$\therefore \overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}=\overrightarrow{a}+2\overrightarrow{d}=\overrightarrow{OA}+\overrightarrow{AH}=\overrightarrow{OH}=\overrightarrow{h}\dots \left(1\right)$

According to given information, $HD=DT$

$⟹\frac{\overrightarrow{b}+\overrightarrow{c}}{2}=\frac{\overrightarrow{h}+\overrightarrow{t}}{2}$

$⟹\frac{\overrightarrow{b}+\overrightarrow{c}}{2}=\frac{\overrightarrow{a}+\overrightarrow{b}+\overrightarrow{c}+\overrightarrow{t}}{2}$

$⟹\overrightarrow{a}=-\overrightarrow{t}$

$⟹\overrightarrow{AT}=2\overrightarrow{a}$

$⟹|\overrightarrow{AT}|=2|\overrightarrow{a}|=2R$

Also, from property of side-length of triangle,

$BC=2RsinA=R$ ($\because \angle A={30}^{\circ }$)

$\therefore AT=2BC$