Too easy for China. Lemma 1: Suppose $ EP $ meets $ BD $ at $ K $. Then $ F,K,Q $ are collinear. Proof: It suffices to show $ FK \perp AD $. Let $ AC \cap BD=J $. We have $ \angle KJC= \angle KPC=90^{\circ} $ so $ KJPC $ is cyclic and $ \angle BKE = \angle JKP=\angle JCP=\angle KBE $ making $ \triangle BEK $ $ E- $ isosceles and thus making $ EF \perp BK $ the perpendicular bisector of $ BK $. Let $ EF \cap BK=T $. Then $ \angle KFE=\angle EFB=\angle ACB=\angle ADB $. If $ FK $ hits $ AD $ at $ Q' $, we immediately get $ \angle Q'DT=\angle Q'FT \implies Q'DTF $ is cyclic and $ \angle FTD=90^{\circ}=\angle FQ'D $ meaning $ Q=Q' $ and $ F,K,Q $ are collinear, Similarly, if $ GR $ cuts $ AC $ at $ L $, then $ F,L,Q $ are collinear. Now, $ LQAR $ is cyclic, which gives $ \angle RQF=\angle BAC $. Since $ KQDP $ is also cyclic, $ \angle PQF=\angle CDB $. But ofcourse $ \angle BAC=\angle CDB $, so $ \angle PQF=\angle RQF $.

My proof doesn't require proving the collinearity that ThirdTimeLucky mentioned(even though it's pretty apparent when drawn). It's clear that $G,P,F,R,E$ are concyclic, since $\angle BEF=\angle BAC=\angle BDC=\angle CGF$, therefore $PF=RF$. This means we only have to prove $Q$ is concyclic with $GPFRE$. Through $E$ we construct a line parallel to $BD$ and it intersects $AD$ at $X$, since $\frac {AX}{DX}=\frac {AE}{BE}=\frac {CF}{BF}=\frac {CG}{GD}$, therefore $GX\parallel AC\Rightarrow FEXG$is a rectangle. Since $\angle FGX=90$, hence $Q$ is concyclic with $GPFRE$ and we are done.

XmL wrote: My proof doesn't require proving the collinearity that ThirdTimeLucky mentioned(even though it's pretty apparent when drawn). It's clear that $G,P,F,R,E$ are concyclic, since $\angle BEF=\angle BAC=\angle BDC=\angle CGF$, therefore $PF=CF$. This means we only have to prove $Q$ is concyclic with $GPFRE$. Through $E$ we construct a line parallel to $BD$ and it intersects $AD$ at $X$, since $\frac {AX}{DX}=\frac {AE}{BE}=\frac {CF}{BF}=\frac {CG}{GD}$, therefore $GX\parallel AC\Rightarrow FEXG$is a rectangle. Since $\angle FGX=90$, hence $Q$ is concyclic with $GPFRE$ and we are done. Don't see $PF=CF$.

elim wrote: XmL wrote: My proof doesn't require proving the collinearity that ThirdTimeLucky mentioned(even though it's pretty apparent when drawn). It's clear that $G,P,F,R,E$ are concyclic, since $\angle BEF=\angle BAC=\angle BDC=\angle CGF$, therefore $PF=CF$. This means we only have to prove $Q$ is concyclic with $GPFRE$. Through $E$ we construct a line parallel to $BD$ and it intersects $AD$ at $X$, since $\frac {AX}{DX}=\frac {AE}{BE}=\frac {CF}{BF}=\frac {CG}{GD}$, therefore $GX\parallel AC\Rightarrow FEXG$is a rectangle. Since $\angle FGX=90$, hence $Q$ is concyclic with $GPFRE$ and we are done. Don't see $PF=CF$. Sorry I meant $PF=RF$

Let $M$ and $N$ be midpoints of $AB$ and $BC$ respectively, $S$ be intersection of $AC$ and $BD$,$M'$ be intersection of $MS$ and $DC$ and $N'$ be intersection of $NS$ and $AD$. Angle chase gives us that $MM'$ is perpendicular to $DC$ and $NN'$ perpendicular to $AD$. Let $X$ be intersection of $EP$ and $FQ$. Triangles $MNS$ and $EFX$ are similar and they're homothetical with center $B$ (because $EF$ parallel to $MN$ , $FX$ parallel to $NS$ and $EX$ parallel to $MS$),so we have $X$ is on $BD$. Analogously,we have that $Y$,intersection of $GR$ and $FQ$,lies on $AC$. Note that quadrilatelars $AQYR$ and $QXPD$ are cyclic,so we have $\angle RQF=\angle SAB=90-\angle ABD$ and $\angle PQF=\angle PDS=90-\angle ABD$,and we are done.

if $FQ \cap BD=X$ notice $FE$ bisects $BX$ and so $EXP$ are collinear. Similarly if $Y=EQ \cap AC$ then $RYG$ are collinear. Notice $RYQA$ and $PXQD$ are cyclic and so $\angle RQY = \angle RAC = \angle XDP = \angle XQP$ and we're done.

It is obvious that $REFPG$ is cyclic.Now from simple angle chasing we obtain $FP=RF$.Now denote $X$ be the point of intersection of $BD$ and $EP$.$XSPC$ is cyclic so we conclude $\triangle EBX$ is isosceles $\implies \triangle FXB$ is isosceles which implies $\angle XFB=\angle ACB$.Notice that $\angle BAC + \angle BDA + \angle DBA +\angle DBC =180$ and that $\angle EAD=\angle DAC +\angle BAC$ and $\angle AEF=180 - \angle BAC$.Now we see that $\angle EAQ + \angle AEF + \angle EFX=270$ so $FX$ is perpendicular to $AD$ which implies $X$ lies on $FQ$.Also let $Y$ be the intersection of $AC$ and $GR$ and then by analogy we have that $Y$ lies on $FQ$.The problem is now trivial since both $AQYR$ and $DQXP$ are cyclic.

Claim: $BD,EP,FQ$ concur. Proof: Let $EP\cap BD=E',BD\cap EF=L.$ Then $$\angle BEF=\angle BAC=\angle BDC=\angle E'DP=\angle E'EL=\angle E'EF$$so it follows that $EL$ is both an altitude and angle bisector of $\triangle BEE',$ and hence $E'$ is the point of $BD$ such that $LE'=BL.$ Similarly, if $FQ\cap BD=F'$ we get a similar conclusion, so $E'=F'$ as desired. We can also get that $RG,FQ,AC$ concur; let $RG\cap FQ\cap AC=X,BD\cap EP\cap FQ=Y.$ Then $ARXQ, QYPD$ cyclic so $$\angle RQX=\angle RAX=\angle RAC=\angle BAC=\angle BDC=\angle YDP=\angle YQP=\angle FQP,$$done.

Consider the following projective transformations from $AD$ to $AB$: 1)$h$ is defind as follows:To a point $K$ on $AD$ we find the intersection of the perpendicular to $AD$ at $K$ and $BC$,then this point we project parallel to $AC$ and finally make orthogonal projection of this point on $CD$. 2)We define $g$ as follows:To a point $K$ on $AD$ we find the intersection of the perpendicular to $AD$ at $K$ and $BC$. Then we project it parallel to $BD$ ,make an orthogonal projection on $AB$,say $H$ and finally we find an intersection of the reflection of the line $HK$ across the perpendicular to $AD$ at $K$ and $AB$. It’s clear that both transformations preserves cross-ratio.Thus,they are projective and they can be uniquely determined by three pairs of points. So we just need to prove the result for three arbitrary cases. 1. Consider the point $Q$ to be the orthogonal projection of the intersection of diagonals on $AD$. Then , using Brahmagupta theorem , we conclude that the intersecrion of the perpendicular is the midpoint of $BC$.So it’s easy to see that if $G$ is the intersection of the diagonals then $AQGR$ and $DQGP$ are cyclic and the conclusion follows. 2.Let $Q$ be the orthogonal projection of $C$ on $AD$ in this case we see that $AQGR$ and $DQGP$ are cyclic( just two opposit angles are $90$. 3. Let $Q$ be the orthogonal projection of $B$ on $AD$. This case is analogical to 2. So we have $h$=$g$ and we are done.

Claim: $\overline{BD},\overline{PE},$ and $\overline{QF}$ are concurrent. Proof. Let $X=\overline{PE}\cap\overline{BD}$ and $Q'=\overline{FX}\cap\overline{AD}.$ It suffices to prove $\overline{FQ'}\perp\overline{AD}.$ Notice $$\measuredangle FEB=\measuredangle CAB=\measuredangle CDB=90-\measuredangle DXP=\measuredangle XEF$$so $\triangle EBF\cong\triangle EXF.$ Hence, $$\measuredangle Q'XD+\measuredangle XDQ'=\measuredangle FXB+\measuredangle BCA=\measuredangle XBF+\measuredangle BFE=90.$$$\blacksquare$ Similarly, let $Y=\overline{FQ}\cap\overline{GR}\cap\overline{AC}.$ Then, $DQXP$ and $ARYQ$ are cyclic, so $$\measuredangle PQX=\measuredangle PDX=\measuredangle CAB=\measuredangle YQR.$$$\square$

I'm actually tweaking from China geo. The main claim is that $E$, $F$, $G$, $P$, $Q$ and $R$ are concyclic. This just follows from, \begin{align*} \angle EFG = \angle ERG = \angle EPG = 90 \end{align*}and from noting that $Q$ lies on this circle with easy complex bash. Namely $f = kb + (1 - k)c$, $e = kb + (1 - k)a$, $g = kd + (1 - k)c$ and then use complex foot to get $q = \frac{1}{2}(a + d + f - ad\overline{f})$ for real $k$. The remaining bash isn't bad, and I am too lazy too type it up. Then angle chase using parallel lines and done.

Let C' is the reflection of C in GF. Then C' lies on AC, because CC' || AC. Then let Q' is the intersection of FC' and AD. $\angle DQ'F = 360 - \angle Q'DC - \angle DCF - \angle DFQ' = 360 - \angle ADB - \angle BDC - \angle DCB - 2\angle CFG = 360 - (\angle CFG + \angle BDC + \angle DCB) - \angle CFG - \angle ADB = 180 - (\angle ADB + \angle CAD) = 90.$ So Q=Q' and C' lies on FQ. Because of the symmetry C' lies on GR. Now let B' is the reflection of B in EF. The B' lies on BD, FQ and EP. Now we have cyclic ARC'Q and DQB'P. $\angle PQF = \angle PQB' = \angle PDB' = \angle CDB = \angle CAB = \angle C'AR = \angle C'QR = \angle FQR$