Hence, all sides must be equal. That is AB=BC=CD=DA
Now AB is produced to points E and F as shown in the figure such that
AE=AB=BF
Hence from above statements, AB=BC=CD=DA=AE=BF
Now, join the opposite ends of the rhombus to make its diagonals.
Let us suppose the diagonals intersect at P.
Since the digonals of a rhombus are always perpendicular, In the ?PAB
x + y + 90 = 180
Or, x + y = 90 .................................................(1)

Now, in ?CBF, sides BC and BF are equal, hence their opposite angles must be also equal.
That is ?BCF = ?CFB

For this triangle ?

CBA is an exterior angle which must be equal to the sum of interior opposite angles.
\That is , ?CBA = ?BCF + ?CFB = 2?CFB
Now ?CBA = ?CBP + ?PBA = x + x = 2x ........ since diagonals bisect the angle of a rhombus.
which means, 2?CFB=2x or ?CFB = x ........................................(2)
Similarly, Applying same set of rules for the ?DAE, we can get
?DEA = y ..............................................................................(3)
From equations 1,2 and 3
we get
?DEA + ?CFB = 90 ...................................................(4)
Let us suppose now that the sides ED and CF meet at O as shown in the figure
Thererfore in ?OEF, we have
?DEA + ?CFB + ?EOF = 180
Putting the value from eqn 4 we get
?EOF = 90
Hence ED and CF are perpendicular

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