Prove that the sum of squares of a sides of a rhombus is equal to the sum of the squares of its diagonals......pls give answer.....u'll surely get a ''thumbs up''

Consider a rhombus ABCD, whose diagonals intersect at O.

We know that diagonals of a rhombus bisect each other at 90Â°, so triangles AOB, AOD, BOC and COD are right angled triangles.

In Î”AOB,

AO^{2} + OB^{2} = AB^{2} (Using Pythagoras theorem)

â‡’ AC^{2} + BD^{2} = 4AB^{2}

Similarly, AC^{2} + BD^{2} = 4BC^{2}

AC^{2} + BD^{2} = 4CD^{2}

AC^{2} + BD^{2} = 4AD^{2}

Adding all these equations, we get

4(AB^{2 }+ BC^{2 }+ CD^{2 }+ AD^{2}) = 4(AC^{2} + BD^{2})

Hence AB^{2 }+ BC^{2 }+ CD^{2 }+ AD^{2 }= AC^{2} + BD^{2}

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