If the internal angle bisectors of $\triangle ABC$ meet its circumcircle at $D$, $E$, $F$, then what is the area of $\triangle DEF$?

If the internal angle bisectors of $\triangle ABC$ meet its circumcircle at $D$, $E$, $F$, then what is the area of $\triangle DEF$?

For the solution of this question in my book, it is written \begin{align} \text{area of}\;\triangle DEF &= \frac{|DE||EF||FD|}{4R} \\[4pt] &= \frac{2R \sin\left(\dfrac{B+C}{2}\right)\cdot 2R \sin\left(\dfrac{C+A}{2}\right)\cdot2R \sin\left(\dfrac{A+B}{2}\right)}{4R} \end{align}

I took this to imply that $$|DE|=2R \left(\dfrac{B+C}{2}\right),\quad |EF|= 2R \sin\left(\dfrac{C+A}{2}\right),\quad |FD|=2R \sin\left(\dfrac{A+B}{2}\right)$$

If this is true, where did they get the result from? Would someone please guide me?

If $D$, $E$, $F$ are the intersections of the circumcircle with the bisectors of $A$, $B$, $C$ respectively, then $$\angle ADE=\angle ABE=\angle ABC/2, \quad \angle ADF=\angle ACF=\angle ACB/2,$$ because they are subtended by the same arcs. Then use the law of sines to get $$EF=2R\sin(\angle EDF)=2R\sin(\angle ABC/2+\angle ACB/2).$$