ignore for now @below, urgh, it's actually quite clear, $FC$ can be any line

Nofancyname wrote: Have I made a typo or ? ...so by easy similar (in fact congruent) triangles containing these sides (also using the parallelogram) , we have $FC$ is a diameter It would be good to show these "easy" arguments, since I don't believe your conclusion is true. $FC$ definitely does not have to be a diameter...

Lemma: Given two parallelograms $ABDC$ and $AEFG$, where $E$ is on $BD$ and $C$ is on $FG$, let $H$ be the point of intersection between $EF$ and $CD$. Show that $AH$ bisects the segment $BG$. Proof (Sketch): Use complex numbers and set $A$ and $C$ to be real numbers and let $B,E$ and $D$ be points of the form $x+i$, and set $H$ to be a random point on $CD$. Now we get $F$ easily and compute $G=A+F-E$, then compute the midpoint to get that it is on $AH$.

$\color{red}\rule{25cm}{0.5pt}$ Now back to the problem. Using the lemma we have that $CT$ bisects $BD$, since both $CQRD$ and $CBSP$ are parallelograms.

Lemma: Given two parallelograms $ABDC$ and $AEFG$, where $E$ is on $BD$ and $C$ is on $FG$, let $H$ be the point of intersection between $EF$ and $CD$. Show that $AH$ bisects the segment $BG$. Proof (Sketch): Use complex numbers and set $A$ and $C$ to be real numbers and let $B,E$ and $D$ be points of the form $x+i$, and set $H$ to be a random point on $CD$. Now we get $F$ easily and compute $G=A+F-E$, then compute the midpoint to get that it is on $AH$.

$\color{red}\rule{25cm}{0.5pt}$ Now back to the problem. Using the lemma we have that $CT$ bisects $BD$, since both $CQRD$ and $CBSP$ are parallelograms.[/quote] Yeah this is what i am looking for during contest. İf we divide problem to two parts, then part 1 will be finding these paralellograms, part to will be provong lemma. İ think lemma's proof is little bit hard, i didn't find anything better.

Easy angle chasing gives $FPCQ$ is a parallelogram, Since $FS = BQ$, clearly $BS = PC$ and $BS // PC$ which means $BSPC$ is a parallelogram. Similarly $DRQC$ is also a parallelogram. $\angle TQC = \angle FDC$ and $\angle TPC = \angle FBC$. Since $FBCD$ is cyclic, $\angle FDC + \angle FBC = 180^{\circ}$. So, $\angle TQC + \angle TPC = 180^{\circ}$, and we get that $TQCP$ is also cyclic. $FPCQ$ and $BSPC$ are both parallelograms, clearly $\angle QCB = \angle FPC$. Since $TQCP$ is cyclic, $\angle TCQ = \angle TPQ$ and hence, adding both of these, $\angle TCB = \angle RPQ$. Similarly $\angle TCD = \angle PQS$. We finish with some law of sinuses Law of sinus on $\triangle RQP$ and $\triangle SPQ$, $\frac{sin \angle PRQ}{sin \angle RPQ} = \frac{PQ}{RQ}$ and $\frac{sin \angle PSQ}{sin \angle PQS} = \frac{PQ}{PS}$. Divide the first by the second to get $$\frac{sin \angle PQS}{sin \angle RPQ} = \frac{PS}{RQ}$$$$\frac{sin\angle TCD}{sin \angle TCB} = \frac{BC}{DC}$$note that by easy angle chasing we can get $\angle PRQ = 180^{\circ} - \angle PSQ$ which means $sin \angle PRQ = sin \angle PSQ$ which is why it can be eliminated. Let $TC$ intersect $BD$ at $M$. By law of sinus on $\triangle MBC$ and $\triangle MDC$ (well known): $$\frac{MD}{MB} = \frac{sin \angle TCD}{sin \angle TCB} \times \frac{sin \angle DBC}{sin \angle BDC}$$ Since $\frac{sin\angle TCD}{sin \angle TCB} = \frac{BC}{DC}$ and by law of sinus on $\triangle BDC$, $\frac{sin \angle DBC}{sin \angle BDC} = \frac{DC}{BC}$. It follows that $\frac{MD}{MB} = 1$ and we are done.

Firstly, since $AF = CD$, we get that, $\angle ABF = \angle CED = \alpha$ . In the same way, we get, that $\angle BAC = \angle FDE = \beta$. $1)\angle AQF = \angle ABQ + \angle BAQ = \alpha + \beta; \angle FPE = \angle PED + \angle PDE = \alpha + \beta .$ $2)\angle BFD = \angle BFC + \angle DFC = \angle BAC + \angle CED = \alpha + \beta .$ So, $\angle AQF = \angle QFP = \angle FPE .$ That means, that $QC || FP$, $QF || PC$ and $FQCP - parallelogram(*) .$ $(*)\implies QC=FP=RD (FR=PD) $, so we get, that $QC=RD$ and $QC||RD \implies QRDC$- parallelogram and $QR||CD(1)$. In the same way, we get, that $SP||CB(2)$. Let $TP\cap CD=X$ and $TQ\cap CB=Y$. $(1),(2)\implies TYCX$- parallelogram $\implies 180^{\circ} - \angle SFR = \angle BCD = \angle QTP = \angle STR \implies SFRT$- сyclic quadrilateral(**). (**)$\implies \angle FSX = \angle FST = \angle TRD = \angle FDX \implies \angle FSX = \angle FDX \implies FSDX$- сyclic quadrilateral and $\triangle FSP \sim \triangle XDP(3) .$ Let $FS=BQ=x, FR=DP=y, CD=a, BC=SP=b$ $$(3)\implies \frac{DX}{x}=\frac{y}{b} \implies xy=DX\cdot b.$$Also, $BY\cdot a = xy.$ As we already know, $TYCX$- parallelogram and $TC$ bisects the diogonal $XY$ . To prove the problem statement it is enough to prove, that $BD||XY\Leftrightarrow \frac{CD}{DX}=\frac{CB}{BY}\Leftrightarrow DX\cdot b=BY\cdot a$. Both of these expressions are equal to $xy$ $\blacksquare .$

whoa, this problem is so cool. I think my solution is pretty new, so I won't hide it. First note that since $BCD$ and $EFA$ are rotations, $BF // CE$. Thus $CQFP$ is a parallelogram. Also note that this gives $\angle{BQC} = \angle{QCP} = \angle{CPD}$, and we also had since the beginning that $\angle{CBF} = 180^\circ - \angle{CDF}$. This gives us that: \begin{align*} \dfrac{CQ}{CB} &= \dfrac{\sin{\angle{CBF}}}{\sin{\angle{CQB}}} \\ &= \dfrac{\sin{\angle{CDF}}}{\sin{\angle{CPD}}} \\ &= \dfrac{CP}{CD} \end{align*}Now we'll look at $S$ and $R$. Note that since $CQ = FP = RD$, we have that $CD // QR$. We also have by ratios that: \begin{align*} \dfrac{BC}{CD} &= \dfrac{CQ}{CP} \\ &= \dfrac{FP}{FQ} \\ &= \dfrac{DR}{BS} \end{align*}Thus, since $\angle{CBF} + \angle{CDF} = 180^\circ$, and $BC \cdot BS = DC \cdot DR$, we now have by the sine area formula that $BCS$ and $DCR$ have the same area. But since $ST$ and $BC$ are parallel, and $RT$ and $CD$ are parallel, we have that $BCT$ and $DCT$ also share this equal area. This is now enough to prove that $CT$ bisects $BD$, and thus we are done!

What does Izho refer to?

Adam_oly wrote: What does Izho refer to? International Zhautykov Olympiad

A quick sketch (haven't fully read the above Solutions); but post $\#6$ mentioned something about projective transformations --- here is a $``\text{one-paragraph}"$ Solution using a collinearity bomb. $\color{green} \rule{25cm}{4pt}$ $\color{green} \textbf{Easy everytime (Just Kidding).}$ Set reference triangle $\triangle BDF$. Now let the midpoints of $BD, BF$ and $FD$ be $M, M_1$ and $M_2$. Reflect $C$ towards $M$ to get $C'$. So it's now left to prove $C,C',T$ collinear. By easy midpoint analysis we get that $CQ \parallel MM_1 \parallel C'S$ and $CP \parallel MM_2 \parallel C'R$. However, Pappus on the lines $l_1 = BF$, $l_2 = DF$ with points $S,Q,P_{\infty} \in \overline{BF}$ and $R,P,P_{\infty} \in \overline{DF}$ yields the desired result. $\blacksquare$ $\blacksquare$ $\blacksquare$ Note: You could also proof the Pappus using Cartesian coordinates (see attachment) without much difficulty.

Note that $AQRF$, $FSPE$ are cyclic. Also, $FRTS$ is cyclic. Let $K=(FRTS)\cap (ABCDEF)$. It is easy to show that $K$ lies on $CT$: $$\measuredangle FKT=\measuredangle FST=\measuredangle FSP=\measuredangle FEP=\measuredangle FEC=\measuredangle FKC.$$ Let $M=CT\cap BD$. \begin{align*} \frac{MD}{MB}&=\frac{\frac{KM\cdot \sin{MKD}}{\sin{KDM}}}{\frac{KM\cdot \sin{MKB}}{\sin{KBM}}} \\&=\frac{\sin{MKD}\cdot \sin{KBM}}{\sin{MKB}\cdot \sin{KDM}} \\&=\frac{\sin{CBD}\cdot \sin{KSR}}{\sin{CDB}\cdot \sin{KRS}} \\&=\frac{\sin{CBD}\cdot KR}{\sin{CDB}\cdot KS} \\&=\frac{\sin{CBD}\cdot RD}{\sin{CDB}\cdot BS} \\&=\frac{\sin{CBD}\cdot FP}{\sin{CDB}\cdot FQ} \\&=\frac{CD\cdot FP}{BC\cdot FQ} \\&=\frac{AF\cdot FP}{EF\cdot FQ} \\&=\frac{BC\cdot FP}{EF\cdot CQ} \\&=\frac{EF\cdot FP}{EF\cdot FP} \\&=1 \end{align*}Done.

It's very cool indeed! Obiviously,$FQCP$,$SPCB$,$QCDR$ are parallelograms and let $O$,$O_1$ and $O_2$ be their centers,respectively. And let $K$ be midpoint of $BD$.Homothety sends $OO_1O_2$ to $FSR$. So this homothey sends $K$ to $K'$,where $K'$ is a point that $SFRK'$ is parallelogram. Since $K'-K-C$ are collinear, we need to prove that $T-K'-C$ are collinear. 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Let $G$ be a point that $AEGF$ is parallelogram and $H=DE\cap BF$. Prove that $H-G-C$ aree collinear. Proof: Let line $GC$ meets $AD$ and $AB$ at $K$, $R$,respectively. And let $BF\cap CG=H'$. We'll prove that $H=H'$. 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Let $DF=a$,$FA=b$,$FG=AE=c$,$EB=d$. Since $EG||DC$ we have $\frac{c+d}{c}=\frac{a+FK}{FK}$ $\implies$ $FK=\frac{ac}{d}$ $\implies$ $KA=b-\frac{ac}{d}$. Since $FG||RA$ we get $\frac{c}{RA}=\frac{\frac{ac}{d}}{b-\frac{ac}{d}}$ $\implies$ $RA=\frac{bd}{a}-c$. Since $FG||RB$ we have $\frac{FH'}{H'B}=\frac{c}{\frac{bd}{a}+d}=\frac{ac}{d(a+b)}$. $\frac{DF}{DA}\cdot \frac{AE}{EB}\cdot \frac{BH'}{H'F}=\frac{a}{a+b}\cdot \frac{c}{d}\cdot \frac{d(a+b)}{ac}=1$. From $\text{Menelaus}$ we get $D-H'-E$ are collinear. $\implies$ $H'=H$. So we are done!

Here is an extremely quick natural solution without adding any additional points in the diagram! The condition is equivalent to $S_{TDC} = S_{TBC}$. From $CD = FA$ and $BC = EF$ we get the isosceles trapezoids $ACDF$ and $CEFB$; in particular $CPFQ$ becomes a parallelogram and from $SB = FQ = CP$ and $DR = PF = CQ$ we also get $CDRQ$ and $BCPS$ as parallelograms. Hence $S_{TDC} = S_{QDC} = S_{QPC} = S_{BPC} = S_{BTC}$, as desired!

Equal lengths induce parallel lines in circles, with $AC \parallel FD$ and $BF \parallel CE$ giving parallelogram $PFQC$. Notice $BS$ is just a translation of $QF$ along line $BF$, so we also get $PSBC$ is a parallelogram. Similarily, along line $DF$, we find $QCDR$ is a parallelogram. Moreover, each of these three parallograms have the same area, and hence $[TBC] = [TCD]$ is simply half of this mutual area. This is sufficient to showing $T$ lies on the $C$-median of $\triangle BCD$. $\blacksquare$