If H is the orthocentre and D is the midpoint of BC of the ∆ABC  and ∠A=30°. Segment HD is produced to T such that HD= DT then  the length AT is equal to

Let the origin of reference be the circumcentre of the triangle Let OA→=a→,OB→=b→,OC→=c→ and OT→=t→ Then |a→|=|b→|=|c→|=R (circumradius)  Again OA→+OB→+OC→=OA→+2OD→                                     =OA→+AH→=OH→Therefore, the P.V. of H is a→+b→+c→ Since D is the midpoint of HT, we have a→+b→+c→+t→2=b→+c→2⇒t→=−a→∴AT→=−2a→ or AT→=|−2a→|=2|a→|=2R.But BC =2R sin A = R; therefore, AT=2BC