Using distance formula, prove the apollonius' theorem that is in `DeltaABC,AB^2+AC^2=2(AD^2+BD^20)` , 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.

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 the figure, A, B, C and D are points on a line in the order. If AC = 5 , BD = 10 and AD = 13, then what is the length of BC ?

In a triangle ABC, B=90^@ and D is the mid-oint of BC then prove that AC^2=AD^2+3 CD^2

In Figure 2, AD _|_ BC . Prove that AB^2 + CD^2 = BD^2 + AC^2 .

In a rhombus ABCD, prove that AC^(2) + BD^(2) = 4AB^(2)

In the figure, AD is perpendicular to BC, prove that AB+AC gt 2AD .

If D, E and F be the middle points of the sides BC,CA and AB of the DeltaABC , then AD+BE+CF is

In DeltaABC, angleB= angleC . Prove the perpendiculars from the mid-point of BC to AB and AC are equal.

In a rectangle ABCD, prove that : AC^(2)+BD^(2)=AB^(2)+BC^(2)+CD^(2)+DA^(2) .