In a , H is the orthocentre and D is the mid point of Segment HD is produced to T such that HD=DT. Then the ratio equals to ;

The correct answer is: 2:1

We should find the ratio of AT to BC using all the given data

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

OA=a

OB=b

OC=c

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

a=b=c=R (circum-radius)

Since D is mid-point of $BC$, d=2b+c​

∴a+b+c=a+2d

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

∴a+b+c=a+2d=OA+AH=OH=h … (1)

According to given information, $HD=DT$

⟹2b+c​=2h+t​

⟹2b+c​=2a+b+c+t​

⟹a=−t

⟹AT=2a

⟹∣AT∣=2∣a∣=2R

Also, from property of side-length of triangle,

$BC=2RsinA=R$ (∵∠A=30∘)

$\therefore AT=2BC$

AT/BC=2:1

Hence 2:1 is the correct option