In a rhombus ABCD show that $4 A B^{2} = A C^{2} + B D^{2}$

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Solution

In rhombus ABCD, AB =BC =CD =DA
We know that diagonals of a rhombus bisect each other perpendicularly.
That is $A C ⊥ B D, ∠ A O B = ∠ B O C = ∠ C O D = ∠ A O D = 90^{\circ}$ and $O A = O C = \frac{A C}{2}, O B = O D = \frac{B D}{2}$
Consider right angled triangle AOB
$A B^{2} = O A^{2} + O B^{2}$ [By Pythagoras theorem]
$\Rightarrow A B^{2} = {(\frac{A C}{2})}^{2} + {(\frac{B D}{2})}^{2} \Rightarrow A B^{2} = \frac{A C^{2}}{4} + \frac{B D^{2}}{4}$
$\Rightarrow 4 A B^{2} = A C^{2} + B D^{2}$

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