In any triangle ABC, show that
`AB^(2)+AC^(2)=2(AD^(2)+BD^(2))`
where, D is the middle point of BC.

In any triangle ABC, prove that AB^2 + AC^2 = 2(AD^2 + BD^2) , where D is the midpoint of BC.

If D(-2, 3), E (4, -3) and F (4, 5) are the mid-points of the sides BC, CA and AB of the sides BC, CA and AB of triangle ABC, then find sqrt((|AG|^(2)+|BG|^(2)-|CG|^(2))) where, G is the centroid of Delta ABC .

In any triangle ABC, prove that cos A= frac{b^2+c^2-a^2}{2bc} .

In Fig. D is a point on side BC of triangleABC such that (BD)/(CD) = (AB)/(AC) . Prove that AD is the bisector of angleBAC .

In fig., D is a point on side BC of triangleABC such that (BD)/(DC)=(AB)/(AC) . Prove that, AD is bisector of angleBAC . .

A, B, C and D are four points in space. Using vector methods, prove that AC^(2)+BD^(2)+AC^(2)+BC^(2)geAB^(2)+CD^(2) what is the implication of the sign of equality.

In any triangle ABC, then by vectors, prove that : c^2=a^2+b^2-2ab cos C .

In Fig if AD bot BC , prove that AB^2 + CD^2 = BD^2 + AC^2 .

Prove that, in any triangle ABC, cos C=(a^2+b^2-c^2)/(2ab) .

In figure, AD is a median of a triangle ABC and AM bot BC. Prove that AC ^(2) + AB ^(2) = 2 AD ^(2) + (1)/(2) BC ^(2).