∆ABC is an isosceles triangle in which AB = AC. D, E and F are the mid-points of the sides BC, AC and AB, respectively. Prove that DE = DF.

To prove DE = DF

Proof In ∆ABC, we have

AB=AC

1/2AB=1/2AC

BF=CE

∠C = ∠B …(ii)

[∵ AB = AC and angles opposite to equal sides are equal]

Now, in ∆ BDF and ∆ CDE, DB = DC

[∵ D is the mid-point of BC]

BF = CE [from Eq. (i)]

and ∠C = ∠B [from Eq. (ii)]

Hence, DF = DE