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Answer:

ABCD is a rhombus AB produced to E and F such that AE=AB=BF.

Construction : Join ED and CF and produce it to meet at G,

To : ED⊥FC prove

proof : AB is produced to points E and F such that

AE=AB=BF __(i)

Also since ABCD is a rhombus

AB=CD=BC=AD __(ii)

Now in ΔBCF,BC=BF [from (i) & (ii)]

1!=2!

3!=1!+2! [exterior angle]

3!=22! __(iii)

Similarly , AE=ED

5!=6!

4!=5!+6!=25! 5!+2!+E!GF=180

o

4!=25!__(iv) E!GF=90

o

by adding (iii) and (iv)

4!+3!=25!+22! ∴4! and 3! are

consective interior angles

∴EG⊥FC Now in ΔEGF Hence it proved

Hope it helps

Answer:

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