Observe that $\angle A F E=\angle A E F=90^{\circ}-A / 2$ and $\angle F D E=\angle A E F=90^{\circ}-A / 2 .$ Again
$\angle E I_{1} F=90^{\circ}+A / 2 .$ Thus
$$
\angle E I_{1} F+\angle F D E=180^{\circ}
$$
Hence $I_{1}$ lies on the incircle. Also
$$
\angle I_{1} F E=(1 / 2) \angle A F E=(1 / 2) \angle A E F=\angle I_{1} E F
$$
Thus $I_{1} E=I_{1} F .$ But then they are equal chords of a circle and so they must subtend equal angles at the circumference. Therefore $\angle I_{1} D F=\angle I_{1} D E$ and so $I_{1} D$ is the internal bisector
of $\angle F D E .$ Similarly we can show that $I_{2} E$ and $I_{3} F$ are internal bisectors of $\angle D E F$ and $\angle D F E$ respectively. Thus the three lines $I_{1} D, I_{2} E, I_{3} F$ are concurrent at the incentre of triangle $D E F$