Question
Mon November 07, 2011 By:

ABCD is a rhombus and AB is produced to E and F such that AE=AB=BF.prove that ED and FC are perpendicular to each other.

Expert Reply
Wed November 09, 2011
 
In the above figure, ABCD is a rhombus.
 
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
 
Thank you
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