Question

In a regular hexagon ABCDEF, prove that ∆BDF is equilateral​

Answer 1

Answer:

Since it is a hexagon, therefore each interior angle will be of measure is 120°

Now ABF is isosceles.

Thus

Similarly we can show that

Thus

.

Hence BDF is equilateral.

Answer 2

to prove:∆BDF is am equilateral

in ∆ABF and ∆EFD

AB=EF(side of a regular hexagon are equal)

AF=ED(side of a regular hexagon are equal)

angle BAC=angle FED(interior angles of a regular hexagon are equal)

therefore,∆ABF congruent to∆EFD(by SAS)

thus, BF=FD(by CPCT)

Similarly,it can be proved FD=DB by proving ∆EFD congruent to ∆CDB

therefore,In ∆BDF,BF=FD=DB

thus ∆BDF is an equilateral triangle.