Question

If D,E and F are the mid-points of the sides BC,CA and AB, respectively of a `DeltaABC` and O is any point, show that
(i) AD+BE+CF=0
(ii) OE+OF+DO=OA

Answer 1

(i) `AD+BE+CF==(d-a)+(e-b)+(f-c)`
`=(d+e+f)-(a+b+c)=0` [usingg eq. (i)]
`implies AD+BE+CF=0`
(ii) Here, `OE+OF+OD=e+f-d`
`=(c+a)/(2)+(a+b)/(2)-(b+c)/(2)=a=OA`
`therefore OE+OF+OD=OA`