Note that $FD \perp NX$, thus $BDFN$ is cyclic. This implies $\angle BND=\angle BFD=\angle BAC=\angle BOD$, where $O$ is the circumcircle of $ABC$. Hence $BOD$ is congruent to $DNB$ and $DN=BO=R$ (note also that $BDOFN$ is cyclic, hence $DN=BO$ are both the diameters of $(BDFN)$)

Let $E\in AC$ such that $F\widehat{D}A=90°$ and $B\widehat{A}C=\alpha$. Then because $DF$ is a middle segment of $\triangle ABC$ then $AC\parallel DF \Rightarrow B\widehat{F}D=\alpha$. See that $A\widehat{F}E=90°-\alpha \Rightarrow N\widehat{F}B=90°-\alpha$. Now, $D\widehat{F}N=\alpha+90°-\alpha=90°$. Note that $D\widehat{F}N+D\widehat{B}N=90°+90°=180°$ then $FNBD$ is cyclic. Let $A\widehat{C}N=\beta$ then as a consequence of inscribed angles $\beta=F\widehat{D}N=A\widehat{C}M$. Let $M$ be such that $C\widehat{A}M=90°, M\in (ABC)$. Note that $D\widehat{F}N=90°=C\widehat{A}M, F\widehat{D}N=\beta=A\widehat{C}M \Rightarrow \triangle DFN \sim CAM$. Note, again, that $DF$ is middle segment then $2DF=AC$ then $2DN=CM=2R \Rightarrow DN=R$.