Center of $(DEF)$ is $M$. Let the feet of altitudes from $M$ to $AD,DC$ be $P,Q$ respectively. $AC\cap BD=N$. Claim: $\angle PNQ=\angle EBF$. Proof: Homothety centered at $D$ with radius $\frac{1}{2}$ gives that $PNQ\sim EBF$. Claim: $\angle PNQ=2\angle ADC$ Proof: \[\angle ADC=\angle CBA=\angle CMA=\angle CMN+\angle NMA=180-\angle NQC+\angle NPA=\angle DQN+\angle NPA\]Since $MNAP$ and $MNQC$ are cyclic. \[2 \angle ADC=\angle PDQ+\angle DQN+\angle NPD=\angle PNQ\] By these claims, we get that $\angle EBF=\angle PNQ=2\angle ADC=\angle EMF\implies E,M,B,F$ are cyclic as desired.$\blacksquare$

Since $ME=MF$, it is enough to prove that $BM$ is external bisector of $\angle EBF$. Since $BM$ is external bisector of $\angle ABC$, it is enough to show that $\angle ABE = \angle CBF$ or $\Delta BAE \sim \Delta BCF$ or (since $\angle BAE = \angle BCF$) that $\frac{AB}{AE}=\frac{BC}{CF} \iff \frac{CD}{AE}=\frac{AD}{CF} \iff AD \cdot AE = CF \cdot CD$, which is true because $AM=MC$ and these expressions are powers of points $A,C$ onto circle $(DEF)$ with center $M$.

Attachments:

This problem seems to have so many nice approaches. Here is another: Let $AC$ and $BD$ intersect at $X$ and let $E'$ and $F'$ be the points along $AD$ and $CD$ such that $\angle BE'D=\angle BF'D=\frac{1}{2}\angle ABC$. We prove $E=E'$ with the proof $F=F'$ being identical. Let $O$ be the circumcenter of $DE'B$. Notice that right triangles $XOB$ and $XMC$ are similar about $X$. By SST triangles $XOM$ and $XBC$ are also similar. Since $OX\perp XB$, we have $OM\perp BC\Rightarrow OM\perp ED$. Thus $ME=MD$ so $E=E'$. We conclude with an angle chase $$\angle EBF=360^{\circ}-\angle EDF-\angle ABC=360^{\circ}-2\angle ABC$$$$\angle EMF=360^{\circ}-\angle EDF-\angle MED-\angle MFD=360^{\circ}-2\angle ABC$$

Attachments:

Let $T,N,K$ be midpoints of $DE,DF,DB$. Note that since $M$ is the center of $EDF$ then $\angle EMF = 360 - 2\angle EDF$. Note that $\angle MKC = 90 = \angle MNC$ and $\angle MKA = 90 = \angle MTA$ so $MKNC$ and $MKAT$ are cyclic. Note that $\angle TKN = 360 - \angle TDN - \angle DTK - \angle DNK = 360 - \angle TDN - \angle AMK - \angle CMK = 360 - \angle TDN - \angle AMC = 360 - \angle ADN - \angle ABC = 360 - 2\angle ADN$ and since $T,N,K$ are midpoints of $DE,DF,DB$ we have $\angle TKN = 360 - 2\angle ADC = \angle EBF = \angle EMF$ so $EMBF$ is cyclic as wanted.