prove that sum of the squares of the diagonal of a rhombus is equal to the sum of the squares of the sides
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In rhombus ABCD, AB = BC = CD = DA
We know that diagonals of a rhombus bisect each other perpendicularly.
That is AC ⊥ BD, ∠AOB=∠BOC=∠COD=∠AOD=90° and
Consider right angled triangle AOB
AB2 = OA2 + OB2 [By Pythagoras theorem]
⇒ 4AB2 = AC2+ BD2
⇒ AB2 + AB2 + AB2 + AB2 = AC2+ BD2
∴ AB2 + BC2 + CD2 + DA2 = AC2+ BD2
Thus the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
hope this helps
We know that diagonals of a rhombus bisect each other perpendicularly.
That is AC ⊥ BD, ∠AOB=∠BOC=∠COD=∠AOD=90° and
Consider right angled triangle AOB
AB2 = OA2 + OB2 [By Pythagoras theorem]
⇒ 4AB2 = AC2+ BD2
⇒ AB2 + AB2 + AB2 + AB2 = AC2+ BD2
∴ AB2 + BC2 + CD2 + DA2 = AC2+ BD2
Thus the sum of the squares of the sides of a rhombus is equal to the sum of the squares of its diagonals.
hope this helps
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