$A B C D$ is a rhombus and its two diagonals meet at $O$ , show that $A B^{2} + B C^{2} + C D^{2} + D A^{2} = 4 A O^{2} + 4 D O^{2}$

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Solution

Consider the given rhombus $A B C D$

Let $A C & B D$ be the diagonals of rhombus $A B C D$

In $Δ A O D$ by phythagoras theorem,

$A B^{2} = A O^{2} + D O^{2}$

By multiplying by $4$ on both side and we get,

$4 A B^{2} = 4 A O^{2} + 4 D O^{2}$

We know that, the all sides of rhombus are equal

Then, $A B = B C = C D = D A$

So, $A B^{2} + B C^{2} + C D^{2} + D A^{2} = 4 A O^{2} + 4 D O^{2}$
Hence proved.

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Similar questions

If $A B C D$ is a rhombus then prove that ${A B}^{2} + {B C}^{2} + {C D}^{2} + {D A}^{2} = {A C}^{2} + {B D}^{2}$ .

A B C D, A B^{2} + B C^{2} + C D^{2} + D A^{2} =

A C

B D

O

A B^{2} + B C^{2} + C D^{2} + D A^{2} = 4 (O A^{2} + O B^{2})

A B^{2} + C D^{2} = B C^{2} + D A^{2}

{AB}^{2} + {BC}^{2} + {CD}^{2} + {DA}^{2} = {AC}^{2} + {BD}^{2} + 4 {PQ}^{2}

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