If D, E, F are the midpoints of the sides BC, CA, AB of a triangle ABC, prove that `bar"AD" + bar"BE" + bar"CF" = bar0`.

#### Solution

Let `bar"a", bar"b", bar"c", bar"d", bar"e", bar"f"` be the position vectors of the points A, B, C, D, E, F respectively.

Since D, E, F are the midpoints of BC, CA, AB respectively, by the midpoint formula

`bar"d" = (bar"b" + bar"c")/2, bar"e" = (bar"c" + bar"a")/2, bar"f" = (bar"a" + bar"b")/2`

∴ `bar"AD" + bar"BE" + bar"CF" = (bar"d" - bar"a") + (bar"e" - bar"b") + (bar"f" - bar"c")`

`= ((bar"b" + bar"c")/2 - bar"a") + ((bar"c" + bar"a")/2 - bar"b") + ((bar"a" + bar"b")/2 - bar"c")`

`= 1/2bar"b" + 1/2bar"c" - bar"a" + 1/2bar"c" + 1/2bar"a" - bar"b" + 1/2bar"a" + 1/2bar"b" - bar"c"`

`= (bar"a" + bar"b" + bar"c") - (bar"a" + bar"b" + bar"c") = bar0`.