$\angle PIB = \angle IBC = \angle IBP$, then $PI = PB$ meaning that $P$ is in the the perpendicular bisector of $IB$. By incenter lemma $D$ is in the perpendicular bisector of $IB$ meaning that $BE = IE$, then $PIEC$ is a parallelogram. Analogously $QIFC$ is a parallelogram. Finally $\triangle ABC$ is not only similar but also homothetic to $\triangle IEF$.

Let $Z$ be the second intersection of $CI$ and $\Gamma$. Since $\measuredangle API = \measuredangle ABC = \measuredangle AZI, (AIPZ)$ are concyclic. Next, $\measuredangle DZC = \measuredangle DAC = \measuredangle PAI = \measuredangle PZC \implies Z,P,D$ collinear. Since $\measuredangle ICE = \measuredangle ACZ = \measuredangle IDE$, $(CIED)$ are concyclic. Then $\measuredangle CEI = \measuredangle CDA = \measuredangle CBA$, so $IE//AB$. Similarly, $IF//AC$ so we are done. $\square$

This fun problem was proposed by Christian Gonzalez from Colombia.

Let $X=AI\cap BC$. Note that $\triangle BDX\sim \triangle ADB$. [asy][asy] import olympiad; import cse5; import geometry; defaultpen(fontsize(10pt)); size(180); pair A=dir(50); pair B=dir(210); pair C=dir(-30); pair D=dir(270); pair I=incenter(A,B,C); pair X=extension(B,C,A,I); pair Pp=B+I-X; pair Qp=C+I-X; pair P=extension(Pp,I,A,B); pair Q=extension(Qp,I,A,C); pair E=extension(B,C,D,P); pair F=extension(B,C,D,Q); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$I$", I, dir(135)); dot("$E$", E, dir(225)); dot("$F$", F, dir(F)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$X$", X, dir(X)); draw(A--B--C--cycle); draw(unitcircle); draw(P--Q); draw(A--D); draw(P--D); draw(Q--D); draw(E--I); draw(F--I); draw(C--D); draw(B--D); [/asy][/asy] Therefore, \[ \frac{DA}{DI}=\frac{DA}{DB}=\frac{DB}{DX}=\frac{DI}{DX}=\frac{DP}{DE}. \]By Thales, this implies that $IE\parallel AB$. In the same way, $IF\parallel AC$. Therefore, triangles $\triangle ABC$ and $\triangle IEF$ have corresponding sides that are parallel, so they are similar, as desired. $\square$