In triangle `A B C ,/_A=30^0,H` is the orthocenter and `D` is the midpoint of `B Cdot` Segment `H D` is produced to `T` such that `H D=D Tdot` The length `A T` is equal to a. `2B C` b. `3B C` c. `4/2B C` d. none of these

Options

A. 2 BC
B. 3 BC
C. `(4)/(3) BC`
D. none of these

Answer

A

Explanation

Let the origin of refrence be the circumcentre of the triangle.
Let `vec(OA) = veca, vec(OB) = vecb, vec(OC) = vecc and vec(OT) = vect`
Then `|veca| = |vecb| = |vecc| = R` (circumradius)
Again `vec(OA) + vec(OB) + vec(OC) = vec(OA) + 2 vec(OD) = vec(OA) + vec(AH) = vec(OH) `

Therefore, the P.V of H is `veca + vecb+vecc`. Since D is the midpoint of HT, we have
`" "(veca +vecb+vecc + vect)/(2) = (vecb+vecc) /(2) rArr vect = - veca`
`therefore vec(AT) = -2veca or vec(AT) = |-2veca| = 2|veca| = 2R`.
But BC = 2R`sin A = R`, therefore,
`AT = 2BC`