Question

ABCD is a rhombus show that AC2 +BD2=4AB2

Answer 1

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⇒ If you draw the diagonals, AC and BD, you can see that they are perpendicular to each other. So they divide the rhombus into 4 congruent right triangles (RT), each with its hypotenuse as one side of the rhombus. AC and BD are each twice the length of each leg of one of those RTs. 

So the sum of the squares of the diagonals, (AC² + BD²), is: 
= 4•(sum of squares of the legs of a single RT), 
and that's just 
= 4•(the square of a side of the rhombus) 
i.e. 
AC² + BD² = 4•AB² 

QED 

EDIT: 
Here's another thing that occurs to me, that is kind of fun: from this answer, you could put the result like this: 

The sum of squares of the diagonals of a rhombus is equal to the sum of squares of its sides.

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