Mathematics
D and E are the mid-points of the sides AB and AC respectively of △ABC. DE is produced to F. To show that CF is equal and parallel to DA, we need an additional information, which is :
DE = EF
AE = EF
∠DAE = ∠EFC
∠ADE = ∠ECF
Mid-point Theorem
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Answer
Assume that, DE = EF
In △ADE and △CFE,
⇒ AE = CE
⇒ DE = EF
⇒ ∠AED = ∠CEF (Vertically opposite angles are equal)
∴ △ADE ≅ △CFE (S.A.S. axiom)
⇒ DA = CF (Corresponding parts of congruent triangles are equal)
⇒ ∠DAE = ∠ECF ..(1) (Corresponding parts of congruent triangles are equal)
⇒ ∠ADE = ∠EFC ..(2) (Corresponding parts of congruent triangles are equal)
Since, ∠DAE and ∠ECF are alternate angles and since they are equal, thus DA // CF.
Thus, if DE = EF then DA is equal and parallel to CF.
Hence, option 1 is the correct option.
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