$A B C D$ is rhombus. Prove that ${A C}^{2} + {B D}^{2} = 4 {A B}^{2}$ .

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Solution

$A B C D$ is a rhombus

$A C$ and $B D$ are diagonals. They bisect perpendicularly at $O$ .

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

In right angled $△ A O B, A^O B = 90^{o}$

According to Pythagoras theorem, we have

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

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

$= \frac{1}{4} {A C}^{2} + \frac{1}{4} {B D}^{2}$

$∴ {A B}^{2} = \frac{1}{4} ({A C}^{2} + {B D}^{2})$

$∴$ ${A C}^{2} + {B D}^{2} = 4 {A B}^{2}$ $[h e n c e p r o v e d]$

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