\[\angle FBD=\angle FBI+\angle IBD=2\angle ABI+2\angle IBC=2\angle ABC\] Obviously the circumcircle of $DEF$ is $I$. \[\angle FED=\frac{1}{2}\angle FED=\frac{1}{2}(180-\angle ABD)\] Hence \[2\angle B+90-\frac{1}{2}\angle B=180\]\[\boxed{\angle B=60.}\]

$ID,IE,IF$ are perpendicular to $BC,CA,AB$ at $L,M,N$ respectively.($IL=LD$ and $IM=ME$,similarly get the other.notice that $\angle IBN=\angle NBF$ and $\angle IBL=\angle LBD$,hence $\angle FBD=2\angle B$,also $\angle IML=\angle IED=\angle C/2$,similarly $\angle IEF=\angle A/2$,but $\angle FBD+\angle FED=2\angle B+\angle A/2+\angle C/2=180$,hence $\angle B=60$.

Since Ab is te perpendicular bisector of FI,we have FB = IB. Now since I is te center of the circle FEDB with radius r where r is the inradius of triangle ABC, we get IB =IF =2r. Hence the triangle FIB is equilateral with ang FIB = 60 and since AB is perpendicular bisector of IF we get ang ABI =30 . Similarly it can be proved that ang DBI is 30 Hence the only possibble value of angel B is 60

We have $ r.cosec\frac {B}{2}=2r\implies sin\frac {B}{2}=\frac {1}{2}\implies \angle \frac {B}{2}=150^0, 30^0 $ , so clearly $ \angle B=60^0 $