# 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

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