Let $X'$ be such that $X'X$ be a diameter of $(ABC)$. b) Inversion wrt the incircle of $ABC$ sends $(ABC)$ to the nine-point circle of $DEF$ and $(AEIF)$ to line $EF$, hence it sends $P$ to $Y$, implying $Y, P, I$ collinear. Notice that since $YEF$ is similar to $YBC$, $\frac{YB}{YC}=\frac{BF}{CE}=\frac{DB}{DC}$, then $YD$ bisects $\angle CYB$, so $Y, D, X'$ are collinear. The fact that $YAQI$ is cyclic implies that $\angle YQI=\angle YAI=\angle YAX'=\angle YXX'$, but $IQ \parallel XX'$, so $X, Q, Y$ are collinear. a) Since $(Y, Q, X), (Y, P, I), (Y, D, X')$ are triples of collinear points and $(PD, IX'), (QD, XX')$ are pairs of parallel lines, homothety centered at $P$ with radius $\frac{YI}{YP}$ sends $PQ$ to $IX$, hence the result immediately follows.