In `DeltaABC, |_(BAC) = 30^@ and H` is the orthocentre, M is the mid point of `BC. HM` is produced to thepoint T such that `HM = MT`. Then `|bar(AT)| = lambda|bar(BC)|` where `lambda` is

In DeltaABC , 'O' is the orthocentre, M is midpoint of BC and angleBAC = 30^(@) . The segment OM is produced to the point T such that OM = MT. Prove that AT=2BC.

In triangle ABC, angleA=30^(@) , 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 lambda BC units, then lambda is equal to -

In triangle ABC,/_A=30^(@),H is the orthocenter 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.(4)/(2)BC d.none of these

If D is the mid point of side BC of a triangle ABC such that vec AB+vec AC=lambdavec AD, write the value of lambda

P is any point on the circumference of the circumcircle of DeltaABC . H is the orthocentre, M is the midpoint of PH and D is the midpoint of BC. Then

P is any point on the circumference of the circumcircle of DeltaABC . H is the orthocentre, M is the midpoint of PH and D is the midpoint of BC. Then

If D is the mid point of side BC of a triangle ABC such that vec A B+ vec A C=lambda\ vec A D , write the value of lambdadot

In DeltaOAB , L is the midpoint of OA and M is a point on OB such that (OM)/(MB)=2 . P is the mid point of LM and the line AP is produced to meet OB at Q. If bar(OA)=bar(a), bar(OB)=bar(b) then find vectors bar(OP) and bar(AP) interms of bar(a) and bar(b) .