Prove using vectors: Medians of a triangle are concurrent.

Explanation

Let A, B and C be vertices of a triangle.
Let D, E and F be the midpoints of the sides BC, AC and AB respectively .Let `vec(OA) = veca, vec(OB) = vecb, vec(OC) = vecc, vec(OD) = vecd, vec(OE) = vece and vec(OF) = vecf` be position vectors of points A, B, C, D, E and F respectively.
Therefore , by midpoint formula,

`vecd = (vecb+vecc)/2 , vece = (veca+vecc)/2 `
and `vecf = (veca+vecb)/2`
` :. 2 vecd = vecb + vecc, 2 vece= veca+ vecc`
and ` 2 vecf = veca+vecb`
` :. 2 vecd + veca = veca+vecb+vecc,`
`2 vece+vecb=veca+vecb+vecc`
` 2 vecf + vecc = veca + vecb+vecc`
Now, ` (2 vecd + veca)/ 3 = (2 vece + vecb)/3`
` = (2 vecf + vecc)/3 = (veca+vecb+vecc)/3 `
Let ` vecg = (veca +vecb+vecc)/3.`
Then , we have `vecg = (veca+vecb +vecc)/3`
` = ((2) vecd + (1) veca)/(2+1)`
`=((2) vece+(1)vecb)/(2 +1)`
` = ((2) vecf +(1) vecc)/(2 +1)`
If G is the point whose position vector is `vec g` , then from the above equation it is clear that the point G lies on the medians AD, BE, CF and it divides them internally in the ratio ` 2 : 1`.
Hence, the medians of a triangle are concurrent .