Prove that in any triangle the sum of squares of any to sides is equal to twice the square of half the third side together with twice the square of themedian

Explanation

Let In triangle ABC, AD is median and AN is perpendicular drawn on BC

Here `AN_|_BC`

In `/_\ABN`

`(AB)^2= (AN)^2 + (BN)^2` ...eq(1)

similarly in `/_\ACN`

`(AC)^2= (AN)^2 + (NC)^2` ...eq(2)

Adding eq(1) and eq(2), we get

`(AB)^2 +(AC)^2= (AN)^2 + (BN)^2 + (AN)^2 + (NC)^2`

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