Q. 155.0( 2 Votes )

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.

Answer :

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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