In the given figure AB =AC and BE ,CF are bisectors of angle B and angle C ,respectively .prove that triangle EBC is congruent to triangle FCB

Question

detail of figure - draw a triangle ABC .then from vertex B draw a angle bisector which meet AC at point E .similarly from vertex C draw angle bisector which again meets AB at point F .PLEASE SIR SOLVE WITH FIGURE AND FULL STEPS .SO THAT I CAN UNDERSTAND CLEARLY .

Answer 1

draw the diagram

AB=AC   given

<ECB=<FBC   base angles are equal in an isosceles triangle

CF bisects <ACB   given

EB bisects <ABC   given

<FCB<EBC   if 2 equal angles are bisected then each half of one angle equals each half of the other angle

CB=CB   identity

therefore triangle ECB is congruent to triangle FCB by ASA

FC=EB   corresponding parts of congruent triangles are congruent

therefore triangle ECB is congruent to triangle FCB by SAS also

 

Answer 2

Label the incenter, I as the intersection of BE and FC.
It is where the angle bisectors meet.


Then by theorem, FI and IE are congruent and IB and IC are congruent,
since they are equal distances to the sides of the triangles from the incenter.
FI = IE and IB = IC.


FC = FI + IC = EI + IB = EB = n <---- let the measure of this side by n


By theorem, the opposite sides of angle bisectors are proportional to
the other sides of the triangle.

so FB / FC = EC / EB

FB / n = EC / n <--- FC = EB = n per step 3


FB = EC <--- multiplies both sides by n; So FB = EC.

Lastly, BC = BC = CB by reflexive property.

The triangles are congruent by SSS.

 

Diagram is available in the resources section with filename EQUAL TRIANGLE MD MERAZ