It was proposed by Greece.

Let $\{D\}=ST\cap BC$. It is easy to see with angle chasing that $AB$ is tangent to $(BCD)$ and $AC$ is tangent to $(BCE)$, hence $AS=\dfrac{AC^2}{AB}$ and $AT=\dfrac{AB^2}{AC}$ so by Menelaus for $S-T-D$ we get $\dfrac{DB}{DC}=\dfrac{AB^2}{AC^2}$, i.e. $A-D-L$ are collinear.

Also barycentric coordinates kills this problem

My solution with barycentric coordinates. We can get easily with angle chasing $AB$ is tangent to $(BCD)$ and $AC$ is tangent to $(BCD).$ Then we know $AS\cdot AB=AC^2,$ and $AT\cdot AC=AB^2.$ We find the coordinate of $S=(1-\frac{b^2}{c^2},\frac{b^2}{c^2},0)$ and $T=(-\frac{c^2}{b^2}+1,0,\frac{c^2}{b^2}).$ And use if the three lines are concurrent the the determinant must be zero.If we compute we get result.

Since ABT and ASC are homothetic (angle chasing), and symmedian AL fits into median in ABT, AST, We get that AK bisects SC. Then By Ceva and Thales on ASC, we are done. K is the intersection of BC and AL.

$CDB \sim BCE \sim ABC$ so $TC $ is tangent to $\odot (BCE)$ and $BS$ is tangent to $\odot (BCD) $ thus by special Reim's we get $CS\parallel TB$ then $TS $ divide $BC$ in the ratio $\frac{TB}{CS}$ .we have $\frac{TB}{BC}=\frac{\sin \angle C}{\sin \angle T}=\frac{\sin \angle C}{\sin \angle B}=\frac{c}{b}$ idem $\frac{CS}{BC}=\frac{\sin \angle (\pi -B)}{\sin \angle S}=\frac{\sin \angle B}{\sin \angle C}=\frac{b}{c}$ hence $\frac{TB}{SC}=\frac{c^2}{b^2}$ therefore $TS$ goes through the foot of the $A$-symmedian so the result follows. RH HAS

This problem was proposed by Vangelis Psychas and Silouanos Brazitikos (me)

silouan wrote: This problem was proposed by Vangelis Psychas and Silouanos Brazitikos (me) Congrats! It's an interesting result!

Let $ST$ and $BC$ intersect at $X$. By Menelaus in $\triangle{AB}$ and points $S, X, T$ we get that $\frac{AS}{SB} \frac{BX}{XC} \frac{CT}{TA}$=$1$. (1) Since $\angle{CBD}=\angle{BCA}$ and $\angle{BCD}=\angle{BAC}$ we get that $\angle{CDB}=180 - \angle{BCD} -\angle{CBD}= \angle{BAC}$, so $AB$ is tangent to the circumcircle of $BCD$. Analogously, we get that $AC$ is tangent to the circumcircle of $SCB$. By power of a point we get that $AB AS=AC^2$ and $AB^=AT AC$. Subtituting in (1) we get that $\frac{XC}{XB}=\frac{AC^3}{AB^3} \frac{CT}{SB}$ (2) But $\frac{TC}{sin(CBT)}=\frac{TC}{sin(B-C)}=\frac{BC}{sin(B}$ (by the sine law). Analogously, $\frac{SB}{sin(B-C)}=\frac{BC}{sin(C)}$. Dividing these two relations we get that $\frac{SB}{TC}=\frac{AC}{AB}$, and substuting in (2) we get that $\frac{BX}{XC}=\frac{AB^2}{AC^2}$, so $AX$ is the symmedian and we are done.

With unnormalised areal coordinates we see: $t_c=a^2 y+b^2 x$ Using the point at infinity on $AC$ as $(1,0,-1)$ we see the line $BD$ has equation: $\begin{vmatrix} x & y & z\\ 1 & 0 & -1\\ 0 & 1 & 0 \end{vmatrix}=0 \implies x+z=0$. Combining this with the equation for $t_c$ we see $D=(a^2,-b^2,-a^2)$. The general circle equation is: $a^2 yz+b^2 xz+c^2 xy+(ux+vy+wz)(x+y+z)=0$ for $u,v,w \in \mathbb{R}$ Circle $BDC$ passes through the points $(0,1,0),(0,0,1),(a^2,-b^2,-a^2)$. The first two points give $v=w=0$ and then putting in $D$ gives $u=-c^2$. Let $S=(x,0,z)$ as $S$ in on line $AC$. Putting this into the equation for the circle we see: $b^2 xz-(c^2 x)(x+z)=0 \implies (b^2-c^2)xz=c^2 x^2 \implies (b^2-c^2)z=c^2 x$ So $S=(b^2-c^2,0,c^2)$ and similarly $T=(c^2-b^2,b^2,0)$. It is well know $AL$ is a symeddian so passes through the symeddian point which has coordinates $(a^2,b^2,c^2)$, Hence $BC \wedge AL=(0,b^2,c^2)$. We prove this point also lies on $ST$: $\begin{vmatrix} 0& b^2 & c^2 \\ b^2-c^2 & 0 & c^2 \\c^2-b^2 & b^2 & 0 \end{vmatrix}=b^2 c^2 (c^2-b^2)+c^2 b^2(b^2-c^2)=0$ So this point also lies on $ST$ and hence $ST,AL,BC$ concur at $(0,b^2,c^2)$

Ratio and angle chasing is enough! Solution: Let $AL\cap BC=N $. Angle chasing gives $\angle SEC=\angle BDC=\angle ATB=\angle ABC$ and $\angle BEC=\angle ABT=\angle ACB $. This implies $\angle TBC=\angle BCS\implies BT||SC$. Now, $AB^2=AT\cdot AC $ and $\frac {AB^2}{AC^2}=\frac {BN}{CN}\implies \frac{BT}{SC}=\frac{BN}{CN}$.

Tumon2001 wrote: Ratio and angle chasing is enough! Solution: $\angle SEC=\angle BDC=\angle ATB=\angle ABC$ Not seeing how $\angle SEC = \angle BDC$ w/o BT || SC, can someone provide insight?

Well $\angle SEC = \angle ABC = 180^\circ - \angle ACD$ since $\angle ACD$ inscribes arc $ABC$. Then $180^\circ - \angle ACD = \angle BDC$ since $BD \parallel AC$.

Let X be the midpoint of CS. (1) Show AC and AB are tangents to the circumcircle of BCE and BDC respectively. (2) AL is the A-symmedian of triangle ABC. (3) Triangle ABC similar to ACS. Consider a transformation that maps the midpoint of BC to X. (4) Show BT is parallel to CS. (5) R(C, S; X, P) is harmonic, where P is the intersection of BT and CS at infinity. Hence the three lines are concurrent.

Here is my solution Note that $BT$ and $SC$ are parallel because they are both anti-parallel to $BC$. Now using the fact that the intersection of a trapezoid is collinear with the midpoints of the parallel sides, and the fact that $AL$ is symmedian, so we are done since $AL$ and the midpoints of the trapezoid, which are anti-parallel, are collinear.

Not as fast as skipiano's but here's another solution Since $BL$ is tangent to $(ABC),$ $\angle CBL=\angle BAC,$ and since $AB\parallel CE,$ $\angle BCE=\angle ABC.$ Combining those two equalities, we get $\angle BEC=\angle ACB,$ so $AC$ is tangent to $(BCE).$ Similarly, $CL$ is tangent to $(ABC),$ so $\angle LCB=\angle BAC,$ and $AC\parallel BD$ implies that $\angle ACB=\angle CBD.$ Hence, $\angle ABC=\angle BDC,$ which is enough to conclude that $AB$ is tangent to $(CBD).$ Let $BC$ meet $AL$ at $X.$ It's well known that $AL$ is the symmedian, so $X$ satisfies the property $\frac{BX}{XC}=\frac{AB^2}{BC^2}.$ Then, by substituting that relation as well as the ones from Power of a Point into Menelaus gives that $S,X,T$ are collinear as desired.

My solution: We know that $CE \parallel AB$ so $\angle{BEC} = 180 - \angle{ABE} = \angle{C}$ and similarly we get $\angle{BDC} = \angle{B}$. It follows by elementary properties of cyclic quadrilaterals that $\angle{ATB} = \angle{B}$ and $\angle{ASC} = \angle{C}$; thus $BT$ and $CS$ are parallel and antiparallel to $BC$. Thus it suffices to show that $AL$ passes through the midpoint of $BT$. Let $K$ be on $(ABC)$ such that $ABKC$ is harmonic; we know that $A, K, L$ collinear. Furthermore, we know that the tangent from $A$ to $(ABC)$ is parallel to $BT$ and $SC$, either by trivial angle-chasing or by elementary antiparallel properties. Thus letting $M$ be the midpoint of $BT$, we see that $(B, T; M, P_{\infty})$ is sent to $(B, C; K, A)$ when we project through $A$ onto $(ABC)$, so we're done.

Oops. I guess I'm late to the party. [asy][asy] size(10cm); import geometry; pair A = dir(60); pair B = dir(120); pair C = dir(230); draw(A--B--C--cycle, orange); draw(circumcircle(A,B,C), red); pair L = (-2+0.3,0.1); draw(L--B^^L--C, deepcyan); pair D = (2*L+C)/3; draw(D--B, heavycyan); draw(circumcircle(B,D,C), blue); pair T = intersectionpoints(circumcircle(B,D,C), A--C)[0]; pair H = C + (-0.1,0); pair E = extension(B,L,C,H); draw(circumcircle(B,C,E), fuchsia); pair U = B + (-10,0); pair S = intersectionpoints(circumcircle(B,C,E), B--U)[1]; draw(B--S); draw(S--T^^A--L, purple+linewidth(1.5)); draw(C--E--S--B^^B--E, heavygreen); draw(B--T--C--S--cycle, magenta+linewidth(1.5)); pair X =extension(B,C,A,L); dot("$A$", A, NE); dot("$B$", B, NW); dot("$C$", C, dir(270)); dot("$D$", D, SW); dot("$L$", L, NW); dot("$S$", S, NW); dot("$T$", T, dir(0)); dot("$X$", X, NW); dot("$E$", E, SW); [/asy][/asy] I'll leave the angle chasing to the reader. Notice that $\angle BCS = \angle ABC - (180^\circ - \angle ABL) = 180^\circ - \angle BCA - \angle LCA = \angle CBT$. Hence, $BT \parallel SC$ and $AB$ is tangent to $(BDC)$. Let $AL \cap BC = X$. It suffices to show that $T,X$ and $S$ are collinear. It's well known that $AL$ is a symmedian so $\tfrac{AB^2}{AC^2} = \tfrac{BX}{XC}$. By PoP, $AB^2 = AT \cdot AC$ so $\tfrac{AT}{AC} = \tfrac{BX}{XC}$. Since, $\triangle ABT$ and $\triangle ASC$ are homothetic, $\tfrac{AT}{AC} = \tfrac{BT}{SC}$. Therefore, $ \tfrac{BX}{XC} = \tfrac{BT}{SC}$ and the result follows.

Popescu wrote: Let $ST$ and $BC$ intersect at $X$. By Menelaus in $\triangle{AB}$ and points $S, X, T$ we get that $\frac{AS}{SB} \frac{BX}{XC} \frac{CT}{TA}$=$1$. (1) Since $\angle{CBD}=\angle{BCA}$ and $\angle{BCD}=\angle{BAC}$ we get that $\angle{CDB}=180 - \angle{BCD} -\angle{CBD}= \angle{BAC}$, so $AB$ is tangent to the circumcircle of $BCD$. Analogously, we get that $AC$ is tangent to the circumcircle of $SCB$. By power of a point we get that $AB AS=AC^2$ and $AB^=AT AC$. Subtituting in (1) we get that $\frac{XC}{XB}=\frac{AC^3}{AB^3} \frac{CT}{SB}$ (2) But $\frac{TC}{sin(CBT)}=\frac{TC}{sin(B-C)}=\frac{BC}{sin(B}$ (by the sine law). Analogously, $\frac{SB}{sin(B-C)}=\frac{BC}{sin(C)}$. Dividing these two relations we get that $\frac{SB}{TC}=\frac{AC}{AB}$, and substuting in (2) we get that $\frac{BX}{XC}=\frac{AB^2}{AC^2}$, so $AX$ is the symmedian and we are done. I think that $\angle{CBD}=\angle{ABC}$

Straightforward angle chasing gives $BT \parallel SC$ and both antiparallel to $BC$. $BC \cap ST={Y}$. Using Ceva's Theorem on $\triangle{ASC}$ we see that $AY$ bisects $BT$ and $SC$ and since they're antiparallel to $BC$, $AY$ is truly the $A-symmedian$ of $\triangle{ABC} \blacksquare$ It seems like the same solution was posted before

Let $Aa$ be the tangent at $A$ of $(ABC)$. - Claim 1: $AB$ tangent to $(BCD)$ (Angle chasing and prove $\widehat{BDC}=\widehat{B}$. Similarly, $AC$ tangent to $(BCE)$ - Claim 2: $BT \parallel Aa$ (Angle chasing and prove $\widehat{aAB}=\widehat{ABT}$). Similarly to $SC$, we have $BT \parallel SC \parallel Aa$ - Claim 3: Let $M=AL \cap BT$. Projecting $A(a,L,B,C)=-1$ on $BT$, we have $M$ is the midpoint of $BT$. Similarly, Let $N=AL \cap SC$, we have $N$ is the midpoint of $SC$ According to a well-known property of trapezoid, we have $SC$ intersect $BC$ at a point in $AN$, i.e. $ST,BC,AL$ concur.

We can angle chase $BT || SC$. So if we let $BC \cap TS = X$, we have that $AX$ is a median of $\triangle ASC$. It's easy to angle chase $\triangle ASC \simeq ACB$, thus $AX$ is a symmedian. But since $AL$ is a symmedian, $L = X$, so we're done.

I don't know why this took such a long time to solve, this was quite easy IMO. BMOSL 2017 G3 wrote: Consider an acute-angled triangle $ABC$ with $AB<AC$ and let $\omega$ be its circumscribed circle. Let $t_B$ and $t_C$ be the tangents to the circle $\omega$ at points $B$ and $C$, respectively, and let $L$ be their intersection. The straight line passing through the point $B$ and parallel to $AC$ intersects $t_C$ in point $D$. The straight line passing through the point $C$ and parallel to $AB$ intersects $t_B$ in point $E$. The circumcircle of the triangle $BDC$ intersects $AC$ in $T$, where $T$ is located between $A$ and $C$. The circumcircle of the triangle $BEC$ intersects the line $AB$ (or its extension) in $S$, where $B$ is located between $S$ and $A$. Prove that $ST$, $AL$, and $BC$ are concurrent. $\text{Vangelis Psychas and Silouanos Brazitikos}$ Let $\omega_1=\odot(BDC)$ and $\omega_2=\odot(BEC)$. Notice that $$\angle ABT=180^\circ-\angle BAT-\angle ATB=180^\circ-\angle BCD-\angle BDC=\angle CBD=\angle TCB$$So $\omega_1$ is tangent to $AB$ at $B$. Similarly we get that $\omega_2$ is tangent to $AC$ at $C$. Now Consider a $\sqrt{bc}$ Inversion with a reflection along the angle bisector of $\angle BAC$. Let this map be $\Psi$. Notice that $\Psi$ swaps $\{\omega_1,\omega_2\}$ and $\{AB,AC\}$. So, $$\Psi(T)=\Psi(AC\cap\omega_1)=\Psi(AC)\cap\Psi(\omega_1)=AB\cap\omega_2=S$$So, $\Psi$ swaps $\{T,S\}$. So, $$AT\cdot AS=AB\cdot AC\implies\frac{AS}{AB}=\frac{AC}{AT}\implies BT\|SC.$$Now let $ST\cap BC=X$ and $AX\cap SC=Y$. Now note that $(SC;Y\infty_{SC})\implies Y$ is the midpoint of $SC$. So if $AX\cap BT=R$, then by a Homothety $(\mathcal H)$ mapping $B$ to $S$ maps $Y$ to $R$. So, $R$ is the midpoint of $BT$. Now by Menelaus Theorem we get that $$1=\frac{CA}{AT}\cdot\frac{TR}{RB}\cdot\frac{BX}{XC}\implies \frac{CA}{AT}=\frac{XC}{XB}\implies \frac{XC}{XB}=\frac{CA^2}{AT\cdot CA}=\frac{AC^2}{AB^2}\implies AX\text{ is the A-Symmedian of }\triangle ABC$$But $L\in AX$. Hence, $ST,BC,AL$ are concurrent at $X$. $\blacksquare$

Cute but Kinda too straightforward Ferid.---. wrote: Consider an acute-angled triangle $ABC$ with $AB<AC$ and let $\omega$ be its circumscribed circle. Let $t_B$ and $t_C$ be the tangents to the circle $\omega$ at points $B$ and $C$, respectively, and let $L$ be their intersection. The straight line passing through the point $B$ and parallel to $AC$ intersects $t_C$ in point $D$. The straight line passing through the point $C$ and parallel to $AB$ intersects $t_B$ in point $E$. The circumcircle of the triangle $BDC$ intersects $AC$ in $T$, where $T$ is located between $A$ and $C$. The circumcircle of the triangle $BEC$ intersects the line $AB$ (or its extension) in $S$, where $B$ is located between $S$ and $A$. Prove that $ST$, $AL$, and $BC$ are concurrent. $\text{Vangelis Psychas and Silouanos Brazitikos}$ Let $X=BC\cap ST$ . Clearly $AB$ is tangent to $\odot{BDC}$,whence $AT=\dfrac{AB^2}{AC}=\dfrac{c^2}{b}$.Thus $\dfrac{CT}{TA}=\dfrac{b^2-c^2}{c^2}$.Similarly $\dfrac{AS}{SB}=\dfrac{b^2-c^2}{c}$.Hence by Menelaus theorm we have \[1=\dfrac{CT}{TA}\cdot \dfrac{AS}{SB}\cdot \dfrac{BX}{XC}=\dfrac{c^2}{b^-c^2}\cdot \dfrac{b^2-c^2}{c^2}\cdot \dfrac{BX}{XC}\implies \dfrac{BX}{XC}=\dfrac{c^2}{b^2}\implies \overline{A-X-L} \text{collinear}\square\]

Since $L$ is the intersection of the tangents at $B$ and $C$, we can see that $AL$ is a symmedian of $\triangle ABC$. Next, observe that $$\angle ASC = 180 - \angle BSC = \angle BDC = 180 - \angle DCA = \frac{1}{2}\overarc{AC} = \angle ABC$$Thus, we have $\triangle ASC \sim \triangle ACB$. Similarly, we also obtain that $\triangle ATB \sim \triangle ABC$. We will proceed using barycentric coordinates. Let our reference triangle be $\triangle ABC$ with $A = (0,1,1), B = (1,0,1), C = (0,0,1)$. Then, since $\triangle ASC \sim \triangle ACB$, we have $$\frac{AC}{AB} = \frac{AS}{AC} \Rightarrow AS = \frac{AC^{2}}{AB} = \frac{b^{2}}{c}$$Thus, since $\frac{AS}{AB} = \frac{b^{2}}{c^{2}}$, we get $S = \left(1-\frac{b^{2}}{c^{2}}, \frac{b^{2}}{c^{2}}, 0\right)$. Similarly, $T = \left(1-\frac{c^{2}}{b^{2}}, 0, \frac{c^{2}}{b^{2}}\right)$ The equation of the line through $S$ and $T$ is $$x(1-\frac{b^{2}}{c^{2}}) - y(\frac{c^{2}}{b^{2}} - 1) + z(1- \frac{b^{2}}{c^{2}}) = 0$$Then, let $X = ST\cap BC$, which implies that $x = 0$. This results in $$-y\left(\frac{c^{2} - b^{2}}{b^{2}}\right) + z\left(\frac{c^{2} - b^{2}}{c^{2}}\right) = 0\Rightarrow -y\left(\frac{1}{b^{2}}\right) + z\left(\frac{1}{c^{2}}\right) = 0 \Rightarrow yc^{2} = zb^{2}$$Thus, the ratio $y:z = b^{2} : c^{2}$. Since any point that lies on the symmedian of $A$ also has the ratio $y:z = b^{2}: c^{2}$, we can conclude that $S$ lies on the symmedian of $A$. Therefore, $ST, BC$, and $AL$ are concurrent.

Solved with @blastoor Define $X = AL \cap BC$. Then $X$ is intersection of $A$ symmedian and $BC$, therefore: $$ \frac{BX}{XC} =\frac{AB^2}{BC^2} $$Claim: $AB$ is tangent to $\odot(TBD)$ Proof: Note that $$\angle CBD =\angle BDT = \angle DTC = \angle C = \angle ABT $$, where we used $\angle ABC = \angle TBD \Longleftrightarrow \angle ABT = \angle CBD$ Analogously we can show that $AC$ is tangent to $\odot(BCE)$. From Pop it follows that: \begin{align*} AB^ 2 = AT \cdot AC \\ \frac{AB^2}{AC^2} = \frac{AT}{AC} \\ \frac{AB^ 2 -AC^2}{AB^2} = \frac{TC}{AT} \end{align*}Similarly we can show that: $$ \frac{AS}{BS} = \frac{AC^2}{AB^2 - AC^2} $$Now using Menelaus theorem in $\triangle ABC$ for points $S,X,T$ we get desired conclusion.

The solution is shorter than the problem let $X=AL\cap BC$, we would prove that $T,X,S$ are collinear $\angle{BDC}=180-\angle{ACD}=\angle{ABC}$ so $AB$ tangents $(BCD)$ similary, $AC$ tangents $(BCE)$ also by angle chasing we can find that $BT\parallel CS$. Because $AL$ is symmedian, $\frac{BX}{XC}=\frac{AB^2}{AC^2}$ $=\frac{AT.AC}{AB.AS}$ $=\frac{AT}{AS}.\frac{AB}{AT}$ $=\frac{AT}{AS}.\frac{BS}{TC}$ by applying Menelaus's theorem, we are done

By angle chasing we have $BDC= B$ .so $(BDC)$ is tangent to $AB$ and power of $A$ gives us $AT =\dfrac{c^2}{b}$ .similarly we have $AS =\dfrac{b^2}{c}$ .now name $F$ the foot of symmedian from $A$ , using sines collinearity from $A$ we're done.

Solution. By angle chasing we get that $AB$ is tangent to $(CBD)$. Now let $X=AL\cap BC$, since $AX$ is the $A$-symmedian we have that $\frac {BX}{CX}=\frac {AB^2}{AC^2}$, also by PoP we have that $AB^2 = AT \cdot AC$. Now using Menelaus' Theorem we get that $S-X-T$ collinear.$\blacksquare$

Here’s an outline Let $AL \cap BC=K$. It is well known that $AK$ is the $A-$ symmedian of $\triangle{ABC}$. Now by some easy angle chasing, we see that $AB^2=AT \times AC$ and $AC^2=AB \times AS$. This tells us that $\frac{KB}{KC}=\frac{AT \times AC}{AB \times AS}$. After this, converse of Menelaus on $\triangle{ABC}$ finishes

Claim1 : $BT || CS$. Proof : $\angle BCS = \angle BCE - \angle SCE = \angle ABC - \angle ACB = \angle BDC - \angle ACB = \angle ATB - \angle ACB = \angle CBT$. Let $AL$ meet $BC$ at $P$ and $ST$ meet $BC$ at $Q$. $AL$ is symmedian in $ABC$ so $\frac{BP}{PC} = (\frac{AB}{AC})^2$. $\frac{BQ}{QC} = \frac{BT}{CS} = \frac{BT.BC}{BC.CS}$ and with sin law we have $ \frac{BT.BC}{BC.CS}$ = $(\frac{AB}{AC})^2$. so $\frac{BQ}{QC} = \frac{BP}{PC}$ and $P$ is $Q$. we're Done.

Let $K=\overline{BC}\cap\overline{AL}.$ Notice $$\measuredangle CBA=\measuredangle CAB+\measuredangle BCA=\measuredangle DCB+\measuredangle CBD=\measuredangle CDB$$so $\overline{AB}$ is tangent to $(BCD)$ at $B.$ Similarly, $\overline{AC}$ is tangent to $(BCE)$ at $C.$ Hence, $$\angle TBA=\angle TCB=\angle CSB$$and $\overline{BT}\parallel\overline{CS}.$ Thus, by PoP, $AT=AB^2/AC$ and $AS=AC^2/AB.$ Also, $CT/SB=AC/AS=AB/AC$ by $\triangle ABT\sim\triangle ASC$ and PoP. By the Properties of a Symmedian Lemma (c), $BK/KC=AB^2/AC^2.$ Menelaus finishes as $$\frac{AS}{SB}\cdot\frac{BK}{KC}\cdot\frac{CT}{TA}=\frac{\frac{AC^2}{AB}}{\frac{AB^2}{AC}}\cdot\frac{AB}{AC}\cdot\frac{AB^2}{AC^2}=1.$$$\square$

Let $K = AL \cap BC$ and $K_1 = BC \cap ST$. It's well-known that $\frac{BK}{KC} = \frac{AB^2}{AC^2}$. Now, observe $$\measuredangle ABD = \measuredangle BAC = \measuredangle BCD$$implying $AB$ is tangent to $(BDCT)$. Analogously, we conclude $AC$ is tangent to $(BSEC)$. Hence, we have $$ATB \overset{-}{\sim} ABC \overset{-}{\sim} ACS$$so $$\measuredangle ABT = - \measuredangle ACB = \measuredangle ASC$$which means $BT \parallel CS$. Now, similarity yields $$\frac{BK_1}{K_1C} = \frac{TB}{CS} = \frac{TB}{BC} \cdot \frac{BC}{CS} = \frac{AB}{AC} \cdot \frac{AB}{AC} = \frac{BK}{KC}$$so $K \equiv K_1$, as required. $\blacksquare$ Remark: Once we determine the angles of $\triangle BCD$ and $\triangle BCE$, this problem becomes trivial.

Note that $AB\cdot AC=AS\cdot AT$ by power of points. Applying $\sqrt{bc}-$inversion on the whole figure, it suffices to show that the $A-$median of $\triangle ABC$ and $(AST)$ meets again on $\omega$. Let the $A-$median meet $\omega$ again at $X\neq A$. Note that $\angle XBS=\angle XCT$ and $$\dfrac{BX}{CX}=\dfrac{AC}{AB}=\dfrac{\dfrac{AC^2}{AB}-AB}{AC-\dfrac{AB^2}{AC}}=\dfrac{AS-AB}{AC-AT}=\dfrac{BS}{CT},$$ implying $\triangle XBS\sim \triangle XCT$. Hence, $A,S,X,T$ are concyclic since $\angle XSA=\angle XSB=\angle XTC$, so we are done.

Let $X=BC \cap ST$ Using the facts $BCES$ and $BTCD$ are isosceles trapezoids , simple angle chasing yields $AB$ is tangent to $(BTCD)$ and $AC$ is tangent to $(BECD).$ Thus $AB^2=AT \cdot AC, AC^2=AB \cdot AS \implies \frac{AB^2}{AC^2}=\frac{AT \cdot AC}{AB \cdot AS}(\clubsuit)$ Since $\angle TBA=\angle TCB=\angle CSB \implies BT || CS \implies \triangle ABT \sim \triangle ASC \implies \frac{BT}{CS}=\frac{AB}{AS}= \frac{AT}{AC}$ In equation $\clubsuit$ if we multiply both sides by $\frac{AB^2}{AC^2}$ we get $$\frac{AB^4}{AC^4}=\frac{AT \cdot AB}{ AS \cdot AC}=\frac{AB^2}{AS^2} \iff \frac{AB^2}{AC^2}=\frac{AB}{AS} \iff \frac{AB^2}{AC^2}=\frac{BT}{CS}$$On the other hand $$\triangle BXT \sim \triangle CXS \implies \frac{BX}{CX}=\frac{BT}{CS} \iff \frac{BX}{CX}=\frac{AB^2}{AC^2}$$ Hence $AX$ is A-symmedian in $\triangle ABC.$ But also it well-known that $AL$ is A-symmedian in $\triangle ABC.$Thus $A,X,L$ are collinear. So we are done

Let $ST \cap BC = X$. An alternative problem formulation is to prove $AX$ is the $A$-symmedian of $\triangle ABC$. First we note that $TBDC$ and $BCES$ are isosceles trapezoids, as they are both cyclic trapezoids. Claim 1: $\triangle ATB \sim \triangle ABC$ \[\angle ABC = 180 - A - C = 180 - \angle BCD - \angle TCB = 180 - \angle TCD = 180 - \angle BTC = \angle ATB.~{\color{blue} \Box}\] Claim 2: $\triangle ATB \sim \triangle ACS$ \[\angle ABT = C = 180 - A - B = 180 - \angle CBE - \angle BCE = \angle BEC = \angle BSC.~{\color{blue} \Box}\] Thus $BTCS$ is a trapezoid whose diagonals meet at $X$. Using similarity ratios, we have \[\frac{BX}{CX} = \frac{BT}{CS} = \frac{AT}{AC} = \frac{AT}{AB} \cdot \frac{AB}{AC} = \left(\frac{AB}{AC}\right)^2,\] and hence $AX$ is the symmedian. $\blacksquare$ [asy][asy] size(250); defaultpen(linewidth(.5)); pair A, B, C, L, D, E, S, T, X; A = dir(120); B = dir(210); C = dir(330); L = extension(B, rotate(90, B)*(0,0), C, rotate(90, C)*(0,0)); D = extension(C, L, B, B+C-A); E = extension(B, L, C, B+C-A); S = B+C+E-2*foot(B, E, C); T = B+C+D-2*foot(C, B, D); X = extension(S, T, B, C); draw(A--B--C--cycle); draw(circumcircle(A, B, C)^^circumcircle(B, C, D)^^circumcircle(B, C, E), gray+linewidth(.4)); draw(B--S--E--C--L--E--L--B--D); draw(B--T--S--C); draw(B--T--C--D--cycle, blue+linewidth(1.5)); draw(B--S--E--C--cycle, red+linewidth(1.5)); dot("$A$", A, NW); dot("$B$", B, W); dot("$C$", C, dir(0)); dot("$L$", L, SW); dot("$D$", D, SE); dot("$E$", E, dir(270)); dot("$S$", S, W); dot("$T$", T, NE); dot("$X$", X, SE); [/asy][/asy]

Let $M$ be the midpoint of $[BC]$ and $BD\cap CE=N, AL\cap SC=P, ST\cap BC=K$. $ACNB$ is a parallelogram so $A,M,N$ are collinear. We know that $\angle PAS=\angle CAM$ because $AP$ is $A-$symmedian and $\angle ASP=\angle C$ so $SPA\sim CMA$ Also $\angle APC=\angle BMA$ and $\angle CAP=\angle MAB$ so $PCA\sim MBA$ So we have $SPCA \sim CMBA$ and this gives us that $P$ is the midpoint of $SC$. $(P\infty,AK\cap SC;S,C)=-1$ and $(P\infty,P;S,C)=-1$ gives us that $P=AK\cap SC$ so $AL$ passes through $ST\cap BC$.

Different solution. Let $BC \cap AL $ at $X$ and $\angle TCB=\angle BDT=\angle CTD=\angle CBD=\alpha$,$\angle TBC=\angle TDC=\beta$,$\angle SBE=x$,$\angle EBD=y$ so, $\angle ABT=180-x-y-\alpha-\beta$ and $\angle ATB=\alpha+\beta$ thus,$\angle BAT=x+y$ which is equal to $\angle EBC=y+\alpha$ from that $x=\alpha$ which gives us $BCED$ tangents to $AC$ also, $\angle BCD=180-2\alpha-\beta=\angle BAC$ that means $BTDC$ tangents to $AB$ that equivalent to $BT$ is parallel to $SC$. We know that Steiner thrm gives us midpoint of $BT,SC$ and $ST \cap BC$ and $A$ are collinear.If we prove $AL$ through midpoint of $BT$ we are done.Call $\angle BAL=\theta$ and symmedian sine lemma tells us (..1)$\frac{\sin(\theta)}{\sin(\alpha)}=\frac{\sin(2\alpha+\beta+\theta)}{\sin(\alpha+\beta)}$ and we have to prove $AX$ is median in $\triangle BAT$ .From median sine to $\triangle BAT$. we need to show $\frac{\sin(\theta)}{\sin(\alpha)}=\frac{\sin(2\alpha+\beta+\theta)}{\sin(\alpha+\beta)}$ which we have found it is true in (..1)

Solution: Let $AC \cap BC =\{X\}, AC \cap SC=\{Y\}$, $\odot (BTCD)=\Gamma_1$, $\odot (BCES)=\Gamma_2$ Now we begin by making the following claim. Claim: $BT \parallel SC$ Proof: $\angle ABT \stackrel{\triangle ABT}{=}180-\angle BTA=\angle BAT=\angle BTC-\angle BAT \equiv \angle BTC-\angle BAC=(\angle BTD+\angle DTC)-\angle BAC \stackrel{\Gamma_1}{=}$ $\stackrel{\Gamma_1}{=} \angle BCD+\angle DBC-\angle BAC \equiv \angle BCL+\angle DBC-\angle BAC \stackrel{\omega}{=} \angle BAC+\angle DBC-\angle BAC=\angle DBC \stackrel{BD \parallel TC}{=} \angle BCT \equiv \angle C \implies \angle ABT=\angle C.$ Also $\angle ASC \equiv BSC \stackrel{\Gamma_2}{=} \angle BEC \stackrel{BS \parallel CE}{=} \angle EBS=180-\angle EBA=180-\angle EBC-\angle CBA \equiv 180-\angle LBC-\angle B \stackrel{\omega}{=} 180-\angle A-\angle B \stackrel{\triangle ABC}{=}$ $ \stackrel{\triangle ABC}{=} \angle C \implies \angle ASC=\angle C$. So $\angle ABT=\angle ASC \implies \boxed{BT \parallel SC}$ $\square$ Claim: $AC^2=AB \cdot AS$ Proof: In the previous claim we found: $\angle ACB=\angle C \equiv \angle ASC \equiv BSC \implies AC-\text{is tangent to} \odot (CBS) \implies \boxed{AC^2=AB \cdot AS}$ $...(1)$ $\square$ Claim: $SY=CY$ Proof: From Ratio Lemma on triangle $\triangle ABC$ we have: $\frac{\sin \angle BAX}{\sin \angle CAX}=\frac{BX}{CX} \cdot \frac{AC}{AB}.$ But since $AX$ is the symedian of triangle $\triangle ABC$ we get that: $\frac{BX}{CX}=\frac{AB^2}{AC^2}.$ Hence $\frac{\sin \angle BAX}{\sin \angle CAX}=\frac{BX}{CX} \cdot \frac{AC}{AB}=\frac{AB^2}{AC^2} \cdot \frac{AC}{AB}=\frac{AB}{AC} \implies \frac{\sin \angle BAX}{\sin \angle CAX}=\frac{AB}{AC} \implies \boxed{\frac{\sin \angle SAY}{\sin \angle CAY}=\frac{AB}{AC}}$ $...(2)$ Again By Ratio Lemma on triangle $\triangle SAC$ we have: $\frac{SY}{CY}=\frac{\sin \angle SAY}{\sin \angle CAY} \cdot \frac{AS}{AC} \stackrel{(2)}{=} \frac{AB}{AC} \cdot \frac{AS}{AC}=\frac{AB \cdot AS}{AC^2} \stackrel{(1)}{=} \frac{AC^2}{AC^2}=1 \implies \boxed{SY=CY}$ $\square$ Claim: $ST,AL$ and $BC$ are concurrent Proof: Since $BT \parallel SC$ from claim 1, we use Thales Theorem to get: $\boxed{\frac{AB}{BS}=\frac{AT}{TC}}$ $...(3)$ So: $\frac{SY}{CY} \cdot \frac{CT}{TA} \cdot \frac{AB}{BS} \stackrel{SY=CY}{=} \frac{CT}{TA} \cdot \frac{AB}{BS} \stackrel{3}{=} 1 \implies \frac{SY}{CY} \cdot \frac{CT}{TA} \cdot \frac{AB}{BS}=1$ which by Ceva's Theorem we get that: $ST,AL$ and $BC$ are concurrent. $\blacksquare$