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

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

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

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 medians of a triangle are concurrent.

Prove that the sum of all vectors drawn from the centre of a regular octagon to its vertices is the zero vector.

Prove that the sum of the three angles of a triangle is 180^0dot

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

Prove using vectors: If two medians of a triangle are equal, then it is isosceles.