In the rhombus $A B C D$ , show that $4 A B^{2} = A C^{2} + B D^{2}$ .

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Solution

In $△ A O B = ∠ O = 90^{\circ}$
$A B^{2} = O A^{2} + O B^{2}$
$= {(\frac{A C}{2})}^{2} + {(\frac{B C}{2})}^{2}$
$= \frac{A C^{2}}{4} + \frac{B D^{2}}{4}$
$A B^{2} = \frac{A C^{2} + B D^{2}}{4}$
$∴ 4 A B^{2} = A C^{2} + B D^{2}$ .

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