The official solution to this problem is so beautiful that I am reproducing it here for visibility. Credit goes to the author, Nikolai Beluhov. Suppose that $ABCD$ has no parallel sides. Let $P = AB \cap CD, Q = AD \cap BC,$ and $R = AC \cap BD$. Furthermore, let the tangents at $A$ and $B$ to the circumcircle of $ABCD$ meet in $S$, those at $B$ and $C$ meet in $T$, those at $C$ and $D$ in $U$, and those at $D$ and $A$ in $V$. [asy][asy] size(12cm); pair A = dir(135); pair B = dir(25); pair C = dir(-30); pair D = dir(-85); pair S = 2*A*B/(A+B); pair T = 2*B*C/(B+C); pair U = 2*C*D/(C+D); pair V = 2*D*A/(D+A); pair P = extension(A, B, C, D); pair Q = extension(A, D, B, C); pair R = extension(A, C, B, D); pair La = extension(B, U, D, T); pair Lc = extension(D, S, B, V); draw(unitcircle); draw(S--T--U--V--cycle, gray); draw(B--U--S--D--T--V--cycle, gray); draw(La--Lc, blue); draw(B--A--D); draw(B--P--D--Q--cycle); string[] names = {"$A$", "$B$", "$C$", "$D$", "$S$", "$T$", "$U$", "$V$", "$P$", "$Q$", "$R$", "$L_a$", "$L_c$"}; pair[] pts = {A, B, C, D, S, T, U, V, P, Q, R, La, Lc}; pair[] labels = {A, dir(45), C, dir(-120), S, T, U, V, P, Q, dir(75), dir(200), dir(135)}; for(int i=0; i<names.length; ++i){ dot(names[i], pts[i], dir(labels[i])); } [/asy][/asy] It is well-known that $R = SU \cap TV$ and that $L_a = BU \cap DT$ and $L_c = BV \cap DS$. By Pappus’s theorem for the hexagon $BUSDTV$ , we see that $R$ lies on $L_aL_c$. Similarly, $R$ lies on $L_bL_d$ and, therefore, $R = L_aL_c \cap L_bL_d$. Analogously, $P = L_aL_b \cap L_cL_d$ and $Q = L_aL_d \cap L_bL_c$. Since the vertices of $\triangle PQR$ are the intersections of the diagonals and opposite sides of $ABCD$, the circumcircle $k$ of $ABCD$ has the property that the polar of any vertex of $\triangle PQR$ with respect to $k$ is the side opposite to that vertex. Analogously, the circumcircle $s$ of $L_aL_bL_cL_d$ has the same property. Given $\triangle PQR$, however, there is exactly one such circle. It follows that $k \equiv s$, and this is a contradiction because $L_aL_bL_cL_d$ lies in the interior of $ABCD$.

Do we have a name for $L_a,L_b,L_c,L_d$???

Rename $L_a$ to $K_A$, etc. since who uses $L$ for symmedian point -_- [asy][asy] size(11cm); defaultpen(fontsize(10pt)); pen pri=red; pen sec=orange; pen tri=fuchsia; pen fil=invisible; pen sfil=invisible; pen tfil=invisible; pair O, A, B, C, D, P, Q, R, EE, F, G, H, KA, KB, KC, KD; O=(0,0); A=dir(150); B=dir(95); C=dir(325); D=dir(205); P=extension(A,B,C,D); Q=extension(A,D,B,C); R=extension(A,C,B,D); EE=2*circumcenter(O,A,B); F=2*circumcenter(O,B,C); G=2*circumcenter(O,C,D); H=2*circumcenter(O,D,A); KA=extension(B,G,D,F); KB=extension(A,G,C,H); KC=extension(B,H,D,EE); KD=extension(A,F,C,EE); draw(H--B--G--C--EE--D--F--A--G,sec+dashed); draw(KB--Q--KA--P--KD--KB,tri); draw(KA--KC,tri); draw(B--Q--A--P--D,pri); draw(EE--G,sec); draw(F--H,sec); filldraw(circumcircle(A,B,C),fil,pri); filldraw(A--B--C--D--cycle,fil,pri); filldraw(EE--F--G--H--cycle,sfil,sec); filldraw(KA--KB--KC--KD--cycle,tfil,tri); dot("$A$",A,A); dot("$B$",B,NE); dot("$C$",C,C); dot("$D$",D,SW); dot("$E$",EE,N); dot("$F$",F,NE); dot("$G$",G,S); dot("$H$",H,W); dot("$P$",P,W); dot("$Q$",Q,N); dot("$R$",R,unit(G-R)); dot("$K_A$",KA,SE); dot("$K_B$",KB,dir(240)); dot("$K_C$",KC,dir(240)); dot("$K_D$",KD,dir(60)); [/asy][/asy] Let $K_A$, $K_B$, $K_C$, and $K_D$ be the symmedian points of $\triangle BCD$, $\triangle CDA$, $\triangle DAB$, and $\triangle ABC$ respectively, and let $E=\overline{AA}\cap\overline{BB}$, $F=\overline{BB}\cap\overline{CC}$, $G=\overline{CC}\cap\overline{DD}$, $H=\overline{DD}\cap\overline{AA}$, $P=\overline{AB}\cap\overline{CD}$, $Q=\overline{AD}\cap\overline{BC}$, and $R=\overline{AC}\cap\overline{BD}$. Assume for the sake of contradiction that $ABCD$ has no two sides parallel (i.e. $P$ and $Q$ are not infinity) but $K_AK_BK_CK_D$ is cyclic. The key claim is this: Claim. $P=\overline{K_AK_B}\cap\overline{K_CK_D}$, $Q=\overline{K_AK_D}\cap\overline{K_BK_C}$, and $R=\overline{K_AK_C}\cap\overline{K_BK_D}$. First proof, using Pappus' Theorem. By definition of symmedian point, $K_A=\overline{BG}\cap\overline{DF}$ and $K_C=\overline{BH}\cap\overline{DE}$. Applying Brianchon's Theorem on hexagons $EBFGDH$ and $EFCGHA$ shows that $R=\overline{EG}\cap\overline{RH}$, so by Pappus' Theorem on $\overline{EBF}$ and $\overline{HDG}$, points $K_A$, $R$, $K_C$ are collinear. By symmetry, $R=\overline{K_AK_C}\cap\overline{K_BK_D}$, and analogous arguments show that $P=\overline{K_AK_B}\cap\overline{K_CK_D}$ and $Q=\overline{K_AK_D}\cap\overline{K_BK_C}$, as desired. $\blacksquare$ Second proof, using pole/polar duality. Take the pole/polar of everything wrt. $(ABCD)$, and let $K=\overline{EH}\cap\overline{CD}$, $N=\overline{GH}\cap\overline{AC}$, and $X=\overline{EG}\cap\overline{CD}$. It is easy to see that $\overline{KN}$ is the polar of $K_B$. However, check that $$-1=(HG;DN)\stackrel K=(E,G;X,\overline{EH}\cap\overline{NK}),$$so $\overline{NK}$ passes through the harmonic conjugate of $X$ wrt. $\overline{EG}$. However, this is a symmetrical condition, so it is immediate that the polar of $K_A$ passes through the same point. Since $P$ is the pole of $\overline{EG}$ by La Hire's, $K_A$, $K_B$, $P$ are collinear, and the claim follows from symmetry. $\blacksquare$ Now, this implies that $(K_AK_BK_CK_D)$ is the polar circle of $\triangle PQR$; however, by definition so is $(ABCD)$. Since the polar circle is unique, $(ABCD)$ and $(K_AK_BK_CK_D)$ coincide, but this is impossible, as the symmedian point of a triangle must lie inside the triangle. $\square$