In triangle ABC ∠ABC = ∠ACB; bisectors of an angle ∠ABC and ∠ACB intersect the side AC and AB at the points E and F respectively. Let’s prove that, FE || BC.

Given.

∠ABC = ∠ACB

BE and CF are the bisectors of ∠ABC and ∠ACB respectively

Formula used.

ASA congruency rule = If 2 angles and one side between them of both triangles are equal then both triangle are congruent.

If 2 triangles are on same base And having equal area then they lies between 2 parallel lines.

As ∠ABC = ∠ACB

Triangle ABC is isosceles triangle

∴ AB = AC

If BE and CF are the bisectors of ∠ABC and ∠ACB

∠ ABE = ∠ CBE and ∠ ACF = ∠ BCF

As ∠ABC = ∠ACB

Then

∠ ABE = ∠ CBE = ∠ ACF = ∠ BCF

In triangle AEB and triangle AFC

AB = AC ∵ Proved above

∠ ABE = ∠ ACF ∵ Proved above

∠ A = ∠ A ∵ common

Triangle AEB ≅ triangle AFC [By ASA property]

If triangle AEB = triangle AFC

Then opening the triangles we get,

In triangle FEB + triangle AFE = triangle FEC + triangle AFE

On subtracting we get

triangle FEB = triangle FEC

In triangle FEB and triangle FEC

triangle FEB = triangle FEC

And they are on same base FE

Hence; they are between parallel lines

FE || BC