Question

24. AABC is an isosceles triangle in which AB=AC, D, E, F are the midpoints of
the sides BC, AC, and AB respectively. Prove that DE=DF.​

Answer 1

Answer:

Given In ∆ABC, AB = AC and D, E and P are the mid-points of the sides BC, AC and AB, respectively.

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