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D and E are the mid-points of the sides AB and AC respectively of \triangle ABC. DE is produced to F. To prove that CF is equal and parallel to DA, we need an additional information which is
(A) \angle DAE = \angle EFC
(B) AE = EF
(C) DE = EF
(D) \angle ADE = \angle ECF.

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Answer:  (C) DE = EF
Solution.
Let us draw the figure according to question.

AE = EC
{E is the mid-point of AC}
Let DE = EF
\angle AED=\angle FEC    {vertically opposite angles}
\therefore \triangle ADE\cong \triangle CFE         {by SAS congruence rule}
\therefore AD=CF       {by CPCT rule}
\therefore \angle ADE=\angle CFE     {by CPCT}
Hence,AD\parallel CF

\ We need DE = EF.
Hence option C is correct.

 

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