If D,E and F are the middle points of the sides BC,CA and AB of the `DeltaABC`, then AD+BE+CF is

Options

A. a zero vector
B. 0
C. a unit vector
D. none of these

Answer

A, B

Explanation

Let D, E and F be the mid-point of the sides of `DeltaABC`.
E and F are mid-point of AC and AB, respectively

So, `EF||BC and EF = (1)/(2)BC`
Thus, AD will also divide EF into equal parts.
Hence, DH is also median of `DeltaDEF`
Similarly BE, median of `DeltaABC`, is also median of `DeltaDEF`.
These two lines meet at G.So the centroids of both triangles are same.
The orthocenter, O of `DeltaDEF` is the point of intersection of the perpendiculars DH and EK drawn from D and E, respectively.

Since, `EF||BC`, DH is perpendicular to BC also.
Similarly EK is perpendicular to AC.
So, orthocenter of `DeltaDEF` and circumcentre of `DeltaABC` is the same points