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

1
by sijinsdaniel
Solved it.. But I think I probably can't write that much
In the morning perhaps

## Answers

• Brainly User
2015-03-07T12:02:57+05:30
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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