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Show that the sum of the sq. of the four sides of a parallelogram is equal to the sum of sq.of two diagonals

$l e t A B C D b e a {| |}^{g m} w i t h d i a g o n a l s A C & B D a s w e k n o w t h a t A D i s t h e m e d i a n o f t r i . (A B C), s o, {(A B)}^{2} + {(A C)}^{2} = 2 {A D}^{2} + \frac{1}{2} {B C}^{2} A S, d i a g o n a l s o f a {| |}^{g m} b i \sec t e a c h o t h e r s . t h e r e f o r e, B O & D O a r e 4 t h e m e d i a n s o f t r i . (A B C) & t r i (A D C) r e s p . s o, {(A B)}^{2} + {(B C)}^{2} = 2 {(B O)}^{2} + \frac{1}{2} {(A C)}^{2} ...................1 {(A D)}^{2} + {(C D)}^{2} = 2 {(D O)}^{2} + \frac{1}{2} {(A C)}^{2} ..................2 a d d i n g (1) & (2), w e g e t {(A B)}^{2} + {(B C)}^{2} + {(C D)}^{2} {+ (A D)}^{2} = 2 ({B O}^{2} + {D O}^{2}) + {(A C)}^{2} {(A B)}^{2} + {(B C)}^{2} + {(C D)}^{2} {+ (A D)}^{2} = 2 (\frac{1}{4} {B D}^{2} + \frac{1}{4} {B D}^{2}) + {A C}^{2} a s, D O = \frac{1}{2} B D \Rightarrow {(A B)}^{2} + {(B C)}^{2} + {(C D)}^{2} {+ (A D)}^{2} = {A C}^{2} + {B D}^{2}$

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