Prove that the sum of three vectors determined by the medians of a triangle directed from the vertices is zero.

The sum of the three vectors determined by the medians of triangle directed from the vertices is

Show that the sum of three vectors determined by the medians of a triangle directed from the vertices is zero.

Point where three medians of a triangle meet.

The sum of these three vectors is :

Three coinitial vectors of magnitudes a, 2a and 3a meet at a point and their directions are along the diagonals if three adjacent faces if a cube. Determined their resultant R. Also prove that the sum of the three vectors determinate by the diagonals of three adjacent faces of a cube passing through the same corner, the vectors being directed from the corner, is twice the vector determined by the diagonal of the cube.

Prove that the sum of the vectors directed from the vertices to the mid-points of opposite sides of a triangle is zero.

Prove that the sum of three altitudes drawn from the vertices to opposite sides of a triangle is less than the sum of three sides. or Prove that the perimeter of a triangle is greater than the sum of three altitudes drawn from the vertices to opposite of a triangle.

Prove that the sum of the three angles of a triangle is 180^(@).

Prove that three times the sum of the squares on the sides of a triangle is equal to four times the sum of the square on the medians of the triangle.

Prove that the locus of the point that moves such that the sum of the squares of its distances from the three vertices of a triangle is constant is a circle.