D and E are the mid-points of the side AB and AC, respectively, of `Delta`ABC. DE is produced to F. To prove that CF is equal and parallel to DA, we need an additional information which is

Options

A. `angleDAE=angleEFC`
B. AE=EF
C. DE=EF
D. `angleADE=angleECF`

Answer

C

Explanation

In `Delta`ADE and `Delta`CFE, suppose DE=EF

Now, `" "` AE=CE `" "` [since, E is the mid-point of AC]
Suppose `" "` DE=EF
and `" "angleAED=angleFEC" "` [vertically opposite angles]
`therefore" "DeltaADE ~= DeltaCFE" "`[by SAS congruence rule]
`therefore" "AD=CF" "`[by CPCT rule] ltBrgt and `" "angleADE=angleCFE" "`[by CPCT]
Hence, `" "AD||CF" "` [since, alternate interior angles are equal]
Therefore, we need an additional information which is DE=EF