In the triangle ABC , B and C are acute angles. The altitude of the triangle drawn from Aintersects BC at D . The bisectors of the angles B and C intersect AD at E and F respectively . If BE = CF , Prove that the triangle ABC is isosceles.
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1 Answers
1we have from the given data
let us assume b>c
then of course B>C (theorem b/sinB=c/sinC)
we have BE as ccosB/cos(B/2)
CF as bcosC/cos(C/2)
as they are equal
ccosB/cos(B/2)=bcosC/cos(C/2)
but as b>c we have
cosBcos(C/2)>cosCcos(B/2)
thus we have
(2cos2(B/2)-1)/cos(B/2) > (2cos2(C/2)-1)/cos(C/2)
simplyfying we get
2cos(B/2)-2cos(C/2)+1/cos(C/2)-1/cos(B/2)>0
thus we get
[cos(B/2)-cos(C/2)](2+1/cosB/2cosC/2)>0
the other term is >0 as B and C are acute
so we have cosB/2>cosC/2 thus B/2<C/2 but then B<C contradiction so we must have b=c
