If D and E are the mid-points of the sides AB and AC respectively of â–³ABC. DE is produced to F. To prove that CF is equal and parallel to DA, we need an additional information which is:
A
△DAE = △EFC
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B
AE=EF
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C
DE=EF
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D
△ADE=△ECF
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Solution
The correct option is CDE=EF Given CF is equal and parallel to DA
D is the midpoint of AB
So, AD =DB= CF= c2
Similarly AE =EC=b2
Let ∠AED=αADE=βandDAE=γ
Since AD is parallel to FC
∠FEC=α∠EFC=βand∠ECF=γ
Applying S.A.S Congruency for triangleADE and CFE, AD=CF, AE=AC
∠DAE=∠ECF
So ΔkDAE andΔkFCE are congruent
So DE=EF
It is given that DE=EF,then by S.S.S. congruency all angles will be equal and CF and parallel to DA