If $A B C D$ is a rhombus then prove that ${4 A B}^{2} = {A C}^{2} + {B D}^{2}$

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Solution

We know that diagonals of a rhombus bisect each other at right angles.

Therefore,

$O A = \frac{1}{2} A C$ , $O B = \frac{1}{2} B D$

Also, $∠ A O B = 90^{o}$

Now, in right $△$ AOB, using the above theorem,

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

$A B^{2} = (\frac{1}{2} A C^{2}) + (\frac{1}{2} B D^{2})$

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

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

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