Solution. Let $I$ be the incenter of $\bigtriangleup ABC$. Line $ID$ is the radical axis of $(BDF)$ and $(CDE)$, so $ID\perp QR$; consequently, $QR\parallel BC$. We can obtain analogous parallelisms for $PR,\ AC$ and $PQ,\ AB$, thus $\bigtriangleup PQR$ and $\bigtriangleup ABC$ are homothetic and hence similar. $\square$

$IA,IB,IC$ - diameters, hence the homothety.

Let $I$ be the incenter of $\triangle ABC$ we have $IE\perp AC$ and $ID\perp BC$ , so $ \angle IEC= \angle IDC=90^\circ$ it means that $IEDC$ is cyclic with diameter $IC$ and since $R$ is the cincuncenter of $\triangle CED$ we'll have $RC=ER=IR=RD$. Similarly $AFIE$ is cyclic so we must have $AP=PE=PI=FP$ therefore $\triangle PIR \cong \triangle PER$ by the $(SSS)$ case so with that we have $IC \perp PR$ and since $IC \perp AC$ it implies $PR \parallel AC$. Doing the same process on the sides sides we will have $QR \parallel BC$ and $QP \parallel AB$, hence they're similar. $\square$