NOTE: BC is not the diameter of the circle, it cannot be assumed here!

Let angle ABC = θ.

Because CD is a tangent to the circle. by the alternate segment theorem, angle ACE = angle ABC = θ.

Because FE is a tangent to the circle, by the alternate segment theorem, angle CAE = angle ABC = θ.

This essentially means that angle CAE = angle ACE = θ.

Angle AEC = 180 - θ - θ = 180 - 2θ
Angle DEF = 180 - (180 - 2θ) = 2θ

so angle DEF = 2 x angle ABC

By analogy,

angle DFE = 2 x angle ACB

ABC = θ
ACE = θ
CAE = θ
DEF = 2θ

Let BCA = x
BAF = x
ABF = x
DFE = 2x

Considering triangle DEF,
EDF = 180 - 2θ - 2x

Considering triangle ABC.
BAC = 180 - θ - x

2 x angle BAC
= 2 (180 - θ - x)
= 360 - 2θ - 2x
= 180 + 180 - 2θ - 2x
= 180 + angle EDF

dont really understand how the analogy is made tho... do we have to do all the workings again to get to this analogy or it can actually be deduced by looking at the diagram?

The fact that the question asks us to “make a similar deduction” means that there has to exist an idea which is very similar to the earlier result.

by the way is the question under the chapter "proofs in plane geometry"?

Yes; this has since been removed from the syllabus