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Use the information in the given figure to prove ;
BD = CF
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In $$ \Delta ABC$$ and $$\Delta EFD $$
$$AB || EF \Rightarrow \angle ABC = \angle EFD $$ (alternate angles )
AC = ED (given )$$ \angle ACB = \angle EFD $$ (given)
$$ \Delta ACB \cong \Delta EFD $$ ( AAS congruence criterion )
Also, $$BC = DF$$ (cpct)
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