Prove that the sum of the squares of the sides of a rhombus is equal to the squares of its diagnol

Prove that the sum of the squares of the sides of a rhombus is equal t

1.	Prove that the sum of the squares of the sides of a rhombus is equal to the squares of its diagnol
Answer» Given: Let, ABCD is a rhombus and since diagonals of a rhombus bisect each other at\xa0{tex} 90 ^ { \\circ }{/tex}To Prove: {tex} \\therefore A B ^ { 2 } + B C ^ { 2 } + C D ^ { 2 } + A D ^ { 2 } = A C ^ { 2 } + B D ^ { 2 }{/tex}Proof :\xa0{tex}\\therefore {/tex}{tex}A O = O C {/tex}\xa0{tex}\\Rightarrow A O ^ { 2 } = O C ^ { 2 }{/tex}{tex}B O = O D {/tex}\xa0{tex}\\Rightarrow B O ^ { 2 } = O D ^ { 2 }{/tex}and\xa0{tex}\\angle A O B = 90 ^ { \\circ }{/tex}\xa0{tex}\\therefore {/tex}\xa0{tex} A B ^ { 2 } = O A ^ { 2 } + B O ^ { 2 }{/tex}Similarly,\xa0{tex} A D ^ { 2 } = x ^ { 2 } + y ^ { 2 } {/tex}{tex}BC ^ { 2 } = x ^ { 2 } + y ^ { 2 } {/tex}{tex}C D ^ { 2 } = x ^ { 2 } + y ^ { 2 } {/tex}{tex} \\therefore A B ^ { 2 } + B C ^ { 2 } + C D ^ { 2 } + D A ^ { 2 } = 4 A O ^ { 2 } + 4 D O ^ { 2 }{/tex}\xa0{tex} = ( 2 A O ) ^ { 2 } + ( 2 D O ) ^ { 2 }{/tex}\xa0{tex} = ( 2 x ) ^ { 2 } + ( 2 y ) ^ { 2 }{/tex}\xa0{tex} \\therefore A B ^ { 2 } + B C ^ { 2 } + C D ^ { 2 } + A D ^ { 2 } = A C ^ { 2 } + B D ^ { 2 }{/tex}\xa0Hence proved.