D and E are the mid-points of the sides AB and AC respectively of ∆ABC. 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
b. AE = EF
c. DE = EF
d. ∠ADE = ∠ECF.

Solution:

In ∆ADE and ∆CFE,

Let us assume DE = EF

As E is the mid-point of AC

AE = CE

If DE = EF

∠AED = ∠FEC [vertically opposite angles]

From the SAS congruence rule

∆ADE ≅ ∆CFE

AD = CF [c.p.c.t]

∠ADE = ∠CFE [c.p.c.t]

We know that alternate angles are equal

AD || CF

Therefore, we need additional information which is DE = EF.

✦ Try This: The figure obtained by joining the mid-points of the adjacent sides 12 cm and 6 cm a. a rhombus, b. a rectangle, c. a square, d. any parallelogram

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 8

NCERT Exemplar Class 9 Maths Exercise 8.1 Problem 14

D and E are the mid-points of the sides AB and AC respectively of ∆ABC. 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, b. AE = EF, c. DE = EF, d. ∠ADE = ∠ECF

Summary:

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

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