triangles 2

we 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