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2014-08-25T20:37:30+05:30

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\frac{AB}{BC} = \frac{EF}{FC} \\ \\ AB * FC = EF * BC \\ \\ \frac{DC}{BC} = \frac{EF}{FB} \\ \\ DC * FB = EF * BC \\ \\ So, \ \ \ AB * FC = DC * FB \\ \\ x FC = y FB \\

Draw a horizontal line at E. then:

 \frac{x - EF} {FB} = \frac{EF }{FC} \\ \\ \frac{x - EF}{ EF} = \frac{FB}{ FC} \\ \\ \frac{x - EF}{ EF} = \frac{x}{y} \\ \\ \frac{ EF (x+y)}{y} = x \\ \\ EF = \frac{x y}{x+y} \\ \\

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  • Brainly User
2014-08-25T22:54:29+05:30
Triangles BFE and BDC are similar.
==> BF/BD = EF/CD
==> BF = ( EF/CD) x BD        ---(1)

Triangles DFE and DBA are similar.
==> FD/BD = EF/AB
==> FD = (EF/AB) x BD        ----(2)

Adding (1) and (2),

      BF + FD = ( EF/CD) x BD + (EF/AB) x BD  = EFxBD[(1/CD) + (1/AB)
As BF + FD = BD, above can be written as

     BD = EFxBD[(1/CD) + (1/AB)
Dividing both sides by BD,
                  1 = EFx( 1/y  + 1/x) = EFx [(x + y)/xy]
                                               
       ==> EF = xy / (x + y)

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