Answer :


Consider ΔABC and AM is the bisector of A and AM bisects the base BC so BM = CM


Extend segment AM to D such that AM = MD and join points C and D to from ΔDMC as shown


To prove ΔABC is isosceles we have to prove that AB = AC


Consider ΔAMB and ΔDMC


AM = MD … construction


AMB = DMC … vertically opposite angles


BM = MC … AM bisects BC given


Hence by SAS test for congruency


ΔAMB ΔDMC


AB = CD … corresponding sides of congruent triangles … (i)


BAM = CDM … corresponding angles of congruent triangles … (a)


BAM = MAC … given AM is angle bisector of A … (b)


Thus using (a) and (b) we can conclude that


CDM = MAC … (c)


Now consider ΔACD


CAD = CDA … from (c)


As the two angles are equal ΔACD is isosceles hence we can say that


AC = CD … (ii)


Now using (i) and (ii) we can conclude that


AB = AC


And hence ΔABC is isosceles triangle


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