It is well known that $D,E,F$ are the circumcentres of$BIC,CIA,AIB$ respectively. So,$ \frac{AR}{sin\frac{C}{2}}= \frac{AF}{sin \frac{A}{2}} =\frac{IF}{sin\frac{A}{2}} $. The angles are easy to be checked. So multipying the ratio of the type $\frac{AR}{IF} $ we get it $1$ .Hence done.

Well, if some one wants to do using complex number then just use the fact that , suppose $T,B,C$ are on a unit circle and $BC$ , line tangent to $T$ meet at $X$ then we've $x=\frac{t(2bc-t(b+c))}{bc-t^2}$ (proof is just by power of a point property)

djb86 wrote: Let $\triangle ABC$ be a triangle with circumcircle $\Gamma$, and let $I$ be the center of the incircle of $\triangle ABC$. The lines $AI$, $BI$ and $CI$ intersect $\Gamma$ in $D \ne A$, $E \ne B$ and $F \ne C$. The tangent lines to $\Gamma$ in $F$, $D$ and $E$ intersect the lines $AI$, $BI$ and $CI$ in $R$, $S$ and $T$, respectively. Prove that \[\vert AR\vert \cdot \vert BS\vert \cdot \vert CT\vert = \vert ID\vert \cdot \vert IE\vert \cdot \vert IF\vert.\] From $D, E, F$ are midpoints arc $BC, CA, AB$ we have $AB\| RF$, $BC\parallel DS$, $ CA\| ET $. $\omega_F , \omega_E , \omega_D $ be circles with centers $F, E,D$ and radiius $FA=FI=FB$, $EA=EC=EI$, $FC=FB=FI$ respectively. Then from powers theorem $ \Rightarrow $ $RF^2-FI^2=RA\cdot RI$ and $RA\cdot RD= RF^2$ $ \Rightarrow $ \[ AR=\frac{FI^2}{DI}.\] Analoguous, $ BS=\frac{DI^2}{EI} $ and $ CT=\frac{EI^2}{FI} $ $ \Longrightarrow $ \[\vert AR\vert \cdot \vert BS\vert \cdot \vert CT\vert = \vert ID\vert \cdot \vert IE\vert \cdot \vert IF\vert.\]