Official Solution: Let $Y$, $Z$, $K$, $L$, $M$ be midpoints of $[BP]$, $[CP]$, $[BC]$, $[CA]$, $[AB]$, respectively. By considering the nine point circles of the triangles $PAB$ and $PAC$, it is easy to see that $(DFX)$ and $(XYM)$ are the same, $(EGX)$ and $(XZL)$ are the same. Let $T$ be the second intersection point of $(XYM)$ and $(XZL)$. By using the parallel lines, simple angle chasing gives: $$\widehat{YTZ}=\widehat{YTX}+\widehat{ZTX}=180^\circ-\widehat{YMX}+180^\circ-\widehat{ZLX}=360^\circ-\widehat{XPY}-\widehat{XPZ}=\widehat{BPC}=\widehat{YKZ}$$This implies $T\in (YKZ)$, so we can conclude that $T$ lies on the nine point circle of the triangle $PBC$. Since $T$ also lies on $BC$, there are two possibilities: $T=K$ or $PT\perp BC$. (i) If $T=K$, angle chasing using parallel lines gives $\widehat{PBA}=\widehat{ABC}-\widehat{PBC}=\widehat{LKC}-\widehat{ZKC}=\widehat{LKZ}$. Hence, by considering the circle $(XLZK)$ we get $\widehat{PCA}=\widehat{LCZ}=\widehat{LXZ}=\widehat{LKZ}$, which implies $\widehat{PBA}=\widehat{PCA}$. (ii) If $PT\perp BC$, we have $|YB|=|YT|$. Then, we get $|MB|=|MT|$ by using $\widehat{MTY}=\widehat{MXY}=\widehat{MBY}$. As a result, $|MA|=|MB|=|MT|$ implies $AT\perp BC$ and the result follows.

Case 1: $DE\parallel FG$ Since $DEFG$ is cyclic, we get $EF=DG$ so $\angle EPF=\angle DPG$. $\Rightarrow \angle DPF=\angle EPG\Rightarrow \angle PFB+\angle PBF=\angle PGC+\angle PCG\Rightarrow \angle PBF=\angle PCG\Rightarrow \angle PBA = \angle PCA$. Case 2: $DE\nparallel FG$ Let's say $DE\cap FG=N, EF\cap DG=M, EG\cap DF=K$ and let the second intersection of the circumcircles of triangles $\Delta DFX$ and $\Delta EGX$ be $H$. When we apply Pascal's Theorem to the hexagon $GAFEPD$, we get $M$ lies on $BC$. By using a known lemma we can see that $H$ lies on $NM$. But we know $H, M \in BC$. Thus, $N$ lies on $BC$ too. Desargues theorem (for the triangles $EGC$ and $DFB$) indicates $ED,GF,CB$ are concurrent $\Leftrightarrow K,P,A$ are collinear. Since $ED,GF,CB$ intersect at $N$, we get $K,P,A$ are collinear. Now, look at the radical axis of the circumcircles of triangles $\Delta DFX$ and $\Delta EGX$. We know the radical axis is $XH$. Since $EDGF$ is cyclic, $FK.KD=EK.KG$. So the point $K$ lies on the radical axis. Thus, $X,K,H$ are collinear $\Rightarrow A,P,H$ are collinear ($A,X,K,P$ are collinear). By Brocard Theorem we get, $XK\perp MN\Rightarrow AP\perp BC$.

Remark 1

Remark 2

Attachments:

mehmetakifyildiz wrote: so we can conclude that $T$ lies on the nine point circle of the triangle $PBC$. It is indeed, the result of the definition of the Poncelet Point on the quadrilateral $ABCP$

BarisKoyuncu wrote: By using a known lemma we can see that $H$ lies on $NM$. Can you tell the lemma?