Let $M$ be the midpoint of arc $\widehat{AC}$ not containing $B$ in $\odot (ABC)$. We claim that $M \in PR$. Animate $D$ on the angle bisector of $\angle ABC$, and note that $$R \mapsto D \mapsto P \mapsto MP \cap AB$$is a projective map. So we just need to check $3$ positions. $D=M$ and when $D$ coincides with the incenter and the $B$-excenter are easily seen to work. Similarly, we have $M \in QS$. Let $MR \cap AC=X,MS \cap AC=Y$. Then, by Shooting Lemma, $MP \cdot MX=MQ \cdot MY$, i.e. $QPXY$ is cyclic. Apply Reim's Theorem to conclude. P.S. There's a typo in the problem statement. $S$ should be on line $BC$.

My solution is using the power ratio lemma on circles $ABC,PQD$ and points $R,S$.

As above, let $M$ be the midpoint of arc $\widehat{AC}$ not containing B in $\odot(ABC)$, so $BD$ passes through $M$. We have $\angle BRD = \angle BAC = \angle BPC = \angle BPD$, so $BRPD$ is cyclic. Similarly, $BCQD$ is cyclic. Also, $\angle DPM = \angle CPM = \angle CBM = \angle ABM = 180 - \angle DPR$ (since $BRPD$ is cyclic). Thus, $M$, $P$, and $R$ are collinear. Similarly, $M$, $Q$, and $S$ are collinear. Then $MP \cdot MR = MD \cdot MB = MQ \cdot MS$, so by converse of power of a point, $RPQS$ is cyclic.

Beautiful problem, love it! matinyousefi wrote: In a convex quadrilateral $ABCD$, $BD$ is the angle bisector of $\angle{ABC}$. The circumcircle of $ABC$ intersects $CD,AD$ in $P,Q$ respectively and the line through $D$ parallel to $AC$ cuts $AB,AC$ in $R,S$ respectively. Prove that point $P,Q,R,S$ lie on a circle. Comment: There is typo in text of exercise where line parallel to $AC$ through $D$ cuts $AB$, $AC$ but it should be that it cuts $AB$ and $BC$. Now here's the solution: If we watch $\triangle ABC$ we get that line $BD$ is angle bisector and that it intersects bigger arc $\widehat AC$ at midpoint $M$. Then we observe that $M, P, R$ and $M, Q, S$ are collinear and that it could finish out solution. But we must prove it. Proof: It is easy to observe that triangles $RBS$ and $ABC$ are homothetic with center $B$ which sends $F$ that is intersetcion of $AC$ and $BD$ to $D$ which is not useful in my solution but I mentioned it to give someone idea for other solution. But useful thing is that $\triangle ABC$ and $\triangle RBS$ are similar $(1)$. Now let's get to angle chase where we will be using $(1)$. $$\angle BRD =\angle BRS =\angle BAC =\angle BPC = \angle BPD (2)$$$$\angle BSD =\angle BSR =\angle BCA =\angle BQA= \angle BQD (3)$$From $(2), (3)$ we get that $BPRD$ and $BSQD$ are cyclic, respectively. Now again by angle chase $$\angle DPM = \angle CPM = \angle CBM = \angle ABM (4)$$Now using $(4)$ and that $BPRD$ is cyclic we get that $$\angle DPM = 180^\circ - \angle DPR (5)$$Similarly by angle chase we get that $$\angle DQM = \angle AQM = \angle ABM =\angle CBM (6)$$Combining $(6)$ and that $BCQD$ is cyclic gives us $$\angle DQM = 180^\circ - \angle DQS (7)$$From $(6)$ and $(7)$ we proved what we desired. Now finally we finish this by obtaining that $$MP*MR = MD*MB = MQ*MS$$Which by reversing power of point implies that $P, Q, R, S$ are concyclic.

We use barycentric coordinates on $\triangle ABC$. In the usual notation, let \[ D = (a : t : c) \]for some $t < 0$; then solving for $P = (a:t:z)$ gives \[ -a^2tz-b^2za-c^2at=0 \implies P = \left( a(b^2+at) : t(b^2+at) : -c^2t \right). \]If we denote by $X$ the image of $P$ under the homothety at $C$ mapping $\triangle BRS$ to $\triangle ABC$, then it follows \begin{align*} X &= \frac{a+t+c}{a+c} \cdot \vec P - \frac{-t}{a+c} \cdot \vec B \\ &= \left( \frac{a(a+t+c)(b^2+at)}{ab^2+a^2t+at^2+b^2t-c^2t} : \frac{t(a+t+c)(b^2+at)}{ab^2+a^2t+at^2+b^2t-c^2t} - t : \frac{a(a+t+c)(-c^2t)}{ab^2+a^2t+at^2+b^2t-c^2t} \right) \\ &= \left( a(a+t+c)(b^2+at) : ct(at+b^2+ct) : -(a+t+c)(c^2t) \right) \end{align*} [asy][asy] size(12cm); pair B = dir(110); pair A = dir(210); pair C = dir(330); pair R = 1.6*A-0.6*B; pair S = extension(B, C, R, R+C-A); pair D = extension(B, dir(270), R, S); pair P = -C+2*foot(origin, C, D); pair Q = -A+2*foot(origin, A, D); pair Y = 0.375*B+0.625*Q; pair X = 0.375*B+0.625*P; filldraw(unitcircle, invisible, blue); filldraw(R--B--S--cycle, invisible, blue); draw(A--C, blue); draw(A--D--C, red); draw(P--B--Q, deepgreen); draw(circumcircle(R, P, Q), orange); draw(circumcircle(C, X, Y), orange); clip(scale(2.2)*unitcircle); dot("$B$", B, dir(B)); dot("$A$", A, dir(180)); dot("$C$", C, dir(0)); dot("$R$", R, dir(270)); dot("$S$", S, dir(270)); dot("$D$", D, dir(D)); dot("$P$", P, dir(P)); dot("$Q$", Q, dir(Q)); dot("$Y$", Y, dir(135)); dot("$X$", X, dir(45)); /* -----------------------------------------------------------------+ | TSQX: by CJ Quines and Evan Chen | | https://github.com/vEnhance/dotfiles/blob/main/py-scripts/tsqx.py | +-------------------------------------------------------------------+ !size(12cm); B = dir 110 A 180 = dir 210 C 0 = dir 330 R 270 = 1.6*A-0.6*B S 270 = extension B C R R+C-A D = extension B (dir 270) R S P = -C+2*foot origin C D Q = -A+2*foot origin A D Y 135 = 0.375*B+0.625*Q X 45 = 0.375*B+0.625*P unitcircle / 0.1 yellow / blue R--B--S--cycle / 0.1 lightcyan / blue A--C / blue A--D--C / red P--B--Q deepgreen circumcircle R P Q / orange circumcircle C X Y / orange !clip(scale(2.2)*unitcircle); */ [/asy][/asy] We now solve for the constant $k$ such that the circle $-a^2yz-b^2zx-c^2xy+(x+y+z)(ky)=0$ passes through $X$. It is given by \begin{align*} k &= \frac{ c^2t(a+t+c) \left[ a^2 \cdot ct(at+b^2+ct) + b^2 \cdot a(a+t+c)(b^2+at) - a(b^2+at) \cdot c(at+b^2+ct) \right] }{ ct(at+b^2+ct) \cdot \left[ a(a+t+c)(b^2+at) + ct(at+b^2+ct) - (a+t+c)(c^2t) \right] } \\ &= \frac{ c^2t(a+t+c) \left[ + b^2 \cdot a(a+t+c)(b^2+at) - ab^2 \cdot c(at+b^2+ct) \right] }{ ct(at+b^2+ct) \cdot \left[ (a+c)\left( a(b^2+at)+ct^2-c^2t \right) + (atb^2+ctb^2)+t^2(a^2-c^2) \right] } \\ &= \frac{ ab^2c^2t(a+t+c) \left[ (a+t+c)(b^2+at) - c(at+b^2+ct) \right] }{ ct(at+b^2+ct)(a+c) \left[ a(b^2+at)+ct^2-c^2t + b^2t + t^2(a-c) \right] } \\ &= \frac{ ab^2c^2t(a+t+c) \left[ ab^2+(a^2+b^2-c^2)t+at^2 \right] }{ ct(at+b^2+ct)(a+c) \left[ ab^2+(a^2+b^2-c^2)t+at^2 \right] } \\ &= \frac{ab^2c(a+c+t)}{(at+ct+b^2)(a+c)}. \end{align*}This expression for $k$ is symmetric in $a$ and $c$. So owing to the symmetry of the problem, the same circle should pass through the analogously defined $Y$. Hence, $XYAC$ is cyclic; therefore $PQRS$ is cyclic too.

Here's a generalization of the problem: Generalized problem wrote: Let $\omega$ be the circumcircle of $\triangle ABC$ and let $D$ be any point on the plane. If $\triangle PQR$ be the projection of $\triangle ABC$ on $\omega$ through $D$ and $PP\cap BC=D'$, then $AA, QR$ and $DD'$ are concurrent. Proof: Suppose $D$ is outside $\omega$. Take a homography fixing $\omega$ and sending $D$ to $\infty$ and the problem is now trivial by symmetry. Because $D$ is "nice", the other cases follow (the condition is some polynomial in $D$, the last argument is due Evan). You can alternatively look at cases about $D$-s position. For the above problem, note this gives that $AA, PQ$ and $RS$ are concurrent. Then look at the PoP of this point to $(ABC)$ and $(ARS)$.