I will let someone else post good solutions.

no $ $ $ $ edit: for completeness, here's the progress I submitted, with a finish by the man who loved only points: Let $P$ be the intersection of $EC$ and $BD$. The similarities imply $\angle ABE = \angle BEC = \angle EDB$ and $\angle AEB = \angle BCE = \angle EBD$. Thus $EC \parallel AB$ and $BD \parallel AE$, so $ABPE$ is a parallelogram. Let $Q$ be the intersection of $(BPC)$ and $(EPD)$ other than $P$, and let $PQ$ hit $EB$ at $M$. Note that since $\angle EBP = \angle ECP$, $(BPC)$ is tangent to line $EB$ at $B$. Similarly $(EPD)$ is tangent to line $EB$ at $E$. Now, $\operatorname{Pow}_{(BPC)} (M) = BM^2$ since $MB$ is tangent to $(BPC)$. Similarly, $\operatorname{Pow}_{(EPD)} (M) = EM^2$. But $PQ$ is the radical axis of $(BPC)$ and $(EPD)$; hence $BM = EM$ and $M$ is the midpoint of $EB$. But since $ABPE$ is a parallelogram, then midpoint of $EB$ is also the midpoint of $AP$. Thus, $A, M, Q, P$ lie on a line. Now we wish to show that $TC \cdot TD = TA^2$. It suffices to show that $T$ lies on the radical axis of $(DPC)$ and $(A)$ (the zero circle centered at $A$), i.e. $\operatorname{Pow}_{(DPC)} (T) - \operatorname{Pow}_{(A)} (T) = 0$. Since $EB$ is tangent to $(BPC)$, we have $EB^2 = EP \cdot EC$, so $\operatorname{Pow}_{(DPC)} (E) = EP \cdot EC = EB^2$. Similarly, $\operatorname{Pow}_{(DPC)} (B) = EB^2$. In addition, we have $\operatorname{Pow}_{(A)} (E) = EA^2$ and $\operatorname{Pow}_{(A)} (B) = BA^2$. Thus, \begin{align*} \operatorname{Pow}_{(DPC)} (E) - \operatorname{Pow}_{(A)} (E) &= EB^2 - EA^2, \\ \operatorname{Pow}_{(DPC)} (B) - \operatorname{Pow}_{(A)} (B) &= EB^2 - BA^2. \end{align*}Now by linearly of power of a point, we have, where lengths are directed, \begin{align*} \operatorname{Pow}_{(DPC)} (T) - \operatorname{Pow}_{(A)} (T) &= \frac{BT}{BE} (EB^2 - EA^2) + \frac{ET}{EB} (EB^2 - BA^2) \\ &= \frac1{1 + ET/TB} \cdot (EB^2 - EA^2) + \frac1{1 + TB/ET} \cdot (EB^2 - BA^2). \end{align*}Let $r = \frac{ET}{TB}$. It suffices to show that $\frac{1+1/r}{1+r} = - \frac{EB^2 - BA^2}{EB^2 - EA^2}$, i.e. $r = -\frac{EB^2 - EA^2}{EB^2 - BA^2}$. But by Menelaus on $\triangle EBP$ and line $D, C, T$, we have \[ \frac{ET}{TB} \cdot \frac{BD}{DP} \cdot \frac{PC}{CE} = -1, \]Now, $\frac{DP}{BD} = -1 - \frac{PB}{BD}$. But $PB = EA$ by parallelogram, and $\frac{BD}{EB} = \frac{EB}{AE}$ by similar triangles, so \[ \frac{DP}{BD} = -1 - \frac{EA}{\frac{EB^2}{AE}} = -1 + \frac{AE^2}{EB^2}. \]Similarly, we have $\frac{PC}{CE} = -1-\frac{EP}{CE} = -1-\frac{AB}{-\frac{EB^2}{AB}} = -1 + \frac{AB^2}{EB^2}$. Thus, \[ r = - \frac{DP}{BD} \cdot \frac{CE}{PC} = -\frac{-1 + \frac{AE^2}{EB^2}}{-1 + \frac{AB^2}{EB^2}} = -\frac{EB^2 - AE^2}{EB^2 - AB^2}.\]We are done. $\blacksquare$

[asy][asy] unitsize(0.4cm); pair A, B, C, D, E, X, Y, T; A=(5,9); B=(0,0); E=(14,0); X=B+E-A; Y=intersectionpoints(A--X,circumcircle(A,B,E))[1]; draw(circumcircle(A,B,E)); draw(circumcircle(B,Y,X)); draw(circumcircle(E,X,Y)); C=(B-A)*abs((B-E)/(A-B))*abs((B-E)/(A-B))+E; D=(E-A)*abs((B-E)/(A-E))*abs((B-E)/(A-E))+B; T=extension(B,E,C,D); draw(A--B--E--A--X--C--T--B--D--C--E); label("$A$", A, NW); label("$B$", B, SW); label("$E$", E, NE); label("$C$", C, SE); label("$D$", D, S); label("$T$", T, NE); label("$X$", X, dir(0)); [/asy][/asy] Let $BD$ and $CE$ intersect at $X$. Then, $$\angle EBD=\angle BCE=\angle BCX$$by similar triangles so $BE$ is tangent to the circumcircle of $BCX$. Similarly, $BE$ is tangent to the circumcircle of $EDX$. We get $$BE^2=BX\times BD$$and $$EB^2=EX\times EC.$$Therefore, we get $$\frac{EC}{EX}=\frac{EB^2}{EX^2}=\frac{EB^2}{AB^2}$$and $$\frac{BD}{BX}=\frac{BE^2}{BX^2}=\frac{BE^2}{AE^2}.$$It suffices to show that $TA^2=TD\times TC$. We will prove this using vectors. Let $a=BE^2$, $b=AE^2$, and $c=AB^2$. Then, $E-C=\frac ac(E-X)=\frac ac(A-B)$, which implies $C=\frac ac(B-A)+E$. Similarly, we get $D=\frac ab(E-A)+B$. Since $-\frac c{b-c}C+\frac b{b-c}D$ lies on $CD$, we get that $$\frac a{b-c}A-\frac a{b-c}B-\frac c{b-c}E-\left(\frac a{b-c}A+\frac b{b-c}B+\frac a{b-c}E\right)=\frac{b-a}{b-c}B+\frac{a-c}{b-c}E$$lies on $CD$. Since this lies on $BE$, we get $T=\frac{b-a}{b-c}B+\frac{a-c}{b-c}E$. Since $TA^2=(T-A)\cdot(T-A)$ and $TC\times TD=(T-C)\cdot(T-D)$, it suffices to show $(T-A)\cdot(T-A)=(T-C)\cdot(T-D)$. We have $T-A=-A+\frac{b-a}{b-c}B+\frac{a-c}{b-c}E$. We also have \begin{align*} T-C&=\frac acA-\frac acB-E+\frac{b-a}{b-c}B+\frac{a-c}{b-c}E\\ &=\frac acA+\frac{bc-ac-ab+ac}{c(b-c)}B+\frac{a-b}{b-c}E\\ &=\frac acA+\frac{b(c-a)}{c(b-c)}B+\frac{a-b}{b-c}E.\end{align*}Similarly, $T-D=\frac abA+\frac{a-c}{c-b}B+\frac{c(b-a)}{b(c-b)}E=\frac cb(T-C)$. Therefore, $(T-C)\cdot(T-D)=\frac cb(T-C)\cdot(T-C)$. Assume without loss of generality the origin is the circumcenter of $ABE$. Then, $A\cdot A=B\cdot B=E\cdot E=R^2$, where $R$ is the circumradius of $ABE$, so $A\cdot B=\frac12(A\cdot A+B\cdot B-(A-B)\cdot(A-B))$, which is equal to $R^2-\frac c2$. Then, we get \begin{align*} &(xA+yB+zE)\cdot(xA+yB+zE)\\ =&x^2A\cdot A+y^2B\cdot B+z^2E\cdot E+2xyA\cdot B+2xzA\cdot E+2yzB\cdot E\\ =&R^2(x^2+y^2+z^2)+2xyR^2-xyc+2xzR^2-xzb+2yzR^2-yza\\ =&R^2(x+y+z)^2-yza-xzb-xyc.\end{align*}Therefore, $(T-A)\cdot(T-A)$ is equal to \begin{align*} &R^2\left(-1+\frac{b-a}{b-c}+\frac{a-c}{b-c}\right)^2-\frac{(b-a)(a-c)a}{(b-c)^2}+\frac{b(a-c)}{b-c}+\frac{c(b-a)}{b-c}\\ =&\frac{b(b-c)(a-c)+c(b-c)(b-a)-a(a-c)(b-a)}{(b-c)^2}\\ =&\frac{bc^2+b^2a-abc-b^2c+cb^2+c^2a-bc^2-abc+a^3-a^2b-a^2c+abc}{(b-c)^2}\\ =&\frac{b^2a+c^2a+a^3-a^2b-a^2c-abc}{(b-c)^2}\\ =&\frac a{(b-c)^2}(a^2+b^2+c^2-ab-bc-ca). \end{align*} Similarly, $(T-C)\cdot(T-C)$ is equal to \begin{align*} &R^2\left(\frac{a(b-c)}{c(b-c)}+\frac{b(c-a)}{c(b-c)}+\frac{c(a-b)}{c(b-c)}\right)^2-\frac{ab(c-a)(a-b)}{c(b-c)^2}-\frac{ab(a-b)}{c(b-c)}-\frac{ab(c-a)}{c(b-c)}\\ =&-\frac{ab}c\left(\frac{(c-a)(a-b)}{(b-c)^2}+\frac{(a-b)(b-c)}{(b-c)^2}+\frac{(c-a)(b-c)}{(b-c)^2}\right)\\ =&-\frac{ab}c\left(\frac{-a^2+ab+ac-bc-b^2+ab+bc-ac-c^2+ac+bc-ab}{(b-c)^2}\right)\\ =&-\frac{ab}c\left(\frac{-a^2-b^2-c^2+ab+bc+ca}{(b-c)^2}\right)\\ =&\frac a{(b-c)^2}\frac bc(a^2+b^2+c^2-ab-bc-ca)\\ =&\frac bc(T-A)\cdot(T-A).\end{align*}Therefore, $(T-C)\cdot(T-D)=\frac cb(T-C)\cdot (T-C)=(T-A)\cdot(T-A)$, so $TC\times TD=TA^2$, which implies $TA$ is tangent to the circumcircle of $ACD$.

Amazing geo, really liked it, but probably it should have been formulated with reference triangle $ABE$. Here goes my solution: Firstly, note that $BC$ and $DE$ are tangent to $(ABE)$ and $BC||AE$ and $DE||AB$ due to the given angles in the similar triangles. Henceforth, we consider $ABE$ as a reference triangle. Next, we introduce the intersection point of $BC$ and $DE$ to be $F$, and let $A'$ be the symmetric point of $A$ wrt $BE$. Note that $ABFE$ is a parallelogram due to the parallel sides and hence $BEA'F$ is an isosceles trapezoid due to $BF=A'E$, $BA'=EF$ and $\angle BFE=\angle BAE=\angle BA'E$ (so it is cyclic quadrilateral with two pairs of equal sides, hence isosceles trapezoid). Since traingles $ABE$ and $BFE$ are congruent, triangles $BFE$, $BCE$ and $BDE$ are similar, hence $BE^2=BC.BF=ED.EF$, so $BC.EA'=ED.BA'$ and thus $BCA'$ and $EDA'$ are similar ($\angle CBA'=\angle DEA'$). Therefore, $DCFA'$ is cyclic. Note that $\angle A'CT=\angle A'FD=\angle A'BT$, so $A'CBT$ is cyclic. Thus $\angle A'TC=\angle A'BC=\angle A'ED$, so $DETA'$ is cyclic. Hence, $\angle A'CT=\angle A'BT=\angle FEB=\angle DA'T$, so $TA'$ is tangent to $CDA'F$. Now, we are ready to finish: we have that $TA^2=TA'^2=TD.TC$, so $TA$ is tangent to $(ACD)$ by power point (we used that $TA=TA'$ due to the symmetry). Edit: This seems to be Solution A of the proposer. Remark: The configuration up to construction of $A, B, C, D, E$ has appeared in Balkan MO 2017 (P2): https://artofproblemsolving.com/community/c6h1441717p8209926

Point circles are hawt!

Debayu ORZ What are point circles?

;=; Reflect $A$ over $BE$ to get $A'.$ Construct point $F$ as the intersection of $BD,CE.$ Let $EA'$ meet $BC$ at $G$ and $BA'$ meet $ED$ at $H.$ Angle chase to get $A',F,D,C,G,H$ all on one circle, then Pascal's $A'A'HDCG$ finishes. $\blacksquare$

Pretty standard ratio solution: Observe the following facts by angle chasing: $C$ and $D$ are isogonal conjugates with respect to $\triangle ABE$. $\overline{BC}$ and $\overline{DE}$ are tangent to $(\triangle ABE)$. $\angle AEC = \angle ABD$. Now, let $\overline{BC}$ and $\overline{DE}$ intersect at $S$. Clearly, $BS = ES$. Now, notice that by Menelaus's Theorem, $$\frac{CT}{TD} \cdot \frac{DE}{SE} \cdot \frac{SB}{BC} = 1 \implies \frac{CT}{TD} = \frac{CB}{ED} = \frac{\frac{CB}{BE}}{\frac{ED}{BE}} = \frac{\frac{AE}{AB}}{\frac{AB}{AE}} = \left(\frac{AE}{AB} \right)^2,$$where the second to last equality follows from the given similar triangles. Now, since $\angle CAE = \angle DAB$ (from the definition of isogonal conjugates) and $\angle AEC = \angle ABD$, triangles $\triangle AEC$ and $\triangle ABD$ are similar. Thus, $$\left(\frac{AC}{AD} \right)^2 = \left(\frac{AE}{AB} \right)^2 = \frac{CT}{TD},$$and we're done.

The given similarities give that $ED$ and $BC$ are tangent to $(ABE)$ , $BD\parallel AE, EC \parallel AB$.Thus , if $p= BD\cap EC$ , then $ABPE$ is a ||gm. Let $Q$ be the reflection of $A$ over $BE$.Thus $PQ\parallel BE$.Also ,$$\angle QBE=\angle EBA =\angle BEP $$Thus $(BEPQ)$ is an isosceles trapezium.Let $D'$ and $C'$ ne points on $BC$ and $ED$ such that $DD' \parallel CC' \parallel BE$.Thus , since its an isosceles trapezium , $(QPDD')$ is cyclic.Similarly , $(QPC'C)$ is cyclic. .Also ,$$\angle PDD' =\angle PBE=\angle BEA=\angle PCD' \implies (PDCD')$$Thus $(QPDC'CD')$ is cyclic.Now , since $B-P-D$ , thus , its refection over the perp bisector of $BE$ gives $D'-Q-E$ and similarly , $C'-Q-B$. Now , pascal on $(QQD'CDC')$ gives that $TQ$ is tanegnt to $(QDC)$ .Thus$$TA^2=TQ^2=TC.TD$$implying the desired result

Let $X$ be the intersection of $BD$ and $CE$. Let $M$ be the miquel point of the self intersecting quadrilateral $BDCE$. This point will magically solve everything. Since $\angle EBX = \angle EBD = \angle ECB = \angle XCB$, we have $(BXC)$ is tangent to $BE$ and analogously $(DXE)$ is too. First a lemma: In a triangle $ABC$, let $P,Q$ be points on $AC, AB$ such that $(APB)$ and $(AQC)$ are tangent to $BC$. Then $(APQ)$ goes through $A'$, the point such that $AA'CB$ is an isosceles trapezoid. To prove this, let $AA'$ meet $CQ$ at $X$. Since $\angle A'XC = 180 - \angle BCQ = 180 - \angle BAC = \angle PAB$ and $\angle APB = \angle ABC = \angle XA'C$ we have triangles $PAB$ and $A'XC$ are similar. Since $AX$ and $BC$ are parallel, the center of spiral similarity sending $AB$ to $XC$ (and hence triangles $PAB$ and $A'XC$ to each other) must be $BA \cap XC = Q$. So $\triangle PQA \sim \triangle A'QX \implies \angle PQA' = \angle AQX = \angle ACB = \angle A'AC$ so we indeed have $(PQAA')$ cyclic, as claimed. Now for the problem, applying the claim to triangle $BXC$ with $P,Q$ being points $C,D$ we see that $M$ is the point such that $BXME$ is an isosceles trapezoid. (Since the miquel lies on $(DXC)$ and $(BXE)$). Because its the miquel point, we also have that $DMBT$ and $EMCT$ are cyclic. So $\angle TMC = \angle TEC = \angle EXM = \angle MDC$ so $TM$ is tangent to $(MDC)$. To finish, since $ABXE$ is a parallelogram (because $\angle XBE = \angle AEB$ and $\angle ABE = \angle XEB$), we have that $A$ is the reflection of $M$ across $BE$. So $A,M$ are symmetric with respect to line $BE$. This means $TA^2 = TM^2 = TC.TD$ by PoP, which implies that $AT$ is tangent to the circumcircle of $\triangle ACD$, as desired.

My problem, but it was not received well at all; the students all hate me now :shrug:

As someone who's not very good at geo, this took me 2 hours Here's my solution. It is easy to see that $AE \parallel BD$ and $AB \parallel EC$. It suffices to show that $\angle TAD = \angle ACD$. By the given similarities, we have \[\frac{DB}{CE} = \frac{EB}{CB} = \frac{BA}{EA} \implies \frac{DB}{BA} = \frac{CE}{EA}\]and $\angle DBA = \angle AEC$ by parallel lines, so $\triangle ABD$ and $\triangle AEC$ are similar. This gives $\angle EAD = \angle BDA = \angle ACE$ Now let $P = AE \cap TD$ and $Q = CE \cap TA$. Notice in $\triangle ABD$ and $\triangle QEP$ that the sides $AB,BD$ are parallel to $QE,EP$ respectively, hence they are homothetic with center $T$. This means \[\triangle QEP \sim \triangle ABD \sim \triangle AEC \implies \angle CAE = \angle EQP\]implying $ACPQ$ is cyclic. Hence, $\angle TAE = \angle ECT$, and finally $\angle TAD = \angle TAE + \angle EAD = \angle ECT + \angle ACE = \angle ACD$, as desired.

One-line solution

Let $A'$ be the reflection of $A$ over $BE$, $BD\cap CE=X$. Since, $A'X || BE$ and $\angle{XA'B}=\angle{ABE}=\angle{BEX}$ so $(BEXA')$ must be cyclic isosceles trapezium. Define $P=EA'\cap BC$ and $Q=BA'\cap DE$. Since, $\angle{A'BE}=\angle{ABE}=\angle{BEX}$ so $BEA'X$ is cyclic. As $DP||A'X$ and $CQ||A'X$, $DPA'X$ and $CQXA'$ are cyclic. Also, $\angle{PCX}=\angle{EBD}=\angle{PDX}$, so $CDPX$ is cyclic. Now, applying pascal's theorem on $A'A'QCDP$, we get that $A'T$ is tangent to $(A'DC)$ Hence, $\overline{TA}^2=\overline{TA'}^2=\overline{TD}\cdot \overline{TC}$ and $TA$ is tangent to $(ADC)$

From $\measuredangle ABE=\measuredangle BEC$ we obtain $AB\parallel CE$ and analogously $AE\parallel BD.$ From $\angle ABD=\angle AEC$ and $$\frac{|BD|}{|CE|}=\frac{|BE|}{|BC|}=\frac{|AB|}{|AE|}$$we obtain $ABD\stackrel{-}{\sim} AEC,$ in particular reflection over bisector of angle $BAE$ followed by projection from $A$ onto $CD$ is the involution by this line, which swaps $(AB\cap CD, AE\cap CD);(C,D).$ By DIT on $B\infty_{AB} E\infty_{AE}$ this involution also swaps $(T,\infty_{CD}),$ so in fact we are done.

hopefully i get points for my poorly explained work :skull:

Actual solution I submitted to this problem Edit: Received a 7. Thanks team leaders for dechipering this and coordinating on my behalf!

From the given three similar triangles we can see that $AB\parallel CE, AE\parallel BD$. Let $I=AE\cap CT, J=CE\cap AT$. Claim: $AD\parallel IJ$. Proof. Since $AB\parallel EJ, BD\parallel EI$ therefore \[ \frac{TJ}{TA}=\frac{TE}{TB}=\frac{TI}{TD}. \]Hence $IJ\parallel AD$.$\blacksquare$ Claim: $A,J,I,C$ are cyclic. Proof. First of all notice that $\triangle ABD\sim\triangle AEC$ because $\angle ABD=\angle AEC$ and \[ \frac{AB}{AE}=\frac{ED}{EB}=\frac{BD}{CE}\implies \frac{AB}{BD}=\frac{AE}{CE} \]Which gives the similarity. Now notice that $\triangle JEI\sim\triangle ABD$ because their corresponding sides are parallel. So $\triangle JEI\sim\triangle ABD\sim\triangle AEC$. Which means $\angle IJC=\angle IAC$. So $A,J,I,C$ are cyclic.$\blacksquare$ Now we have \begin{align*} TI\cdot TA&=TJ\cdot TD\tag{\text{from the first claim}}\\ TI\cdot TC&=TJ\cdot TA\tag{\text{from the second claim}} \end{align*}Combining these two equations we get $TA^2=TC\cdot TD$. And we are done.

Let $A'$ be the image of reflection of $A$ across line $BE$. Then $\square EPA'B$ is cyclic where $P$ is the intersection of $BD$ and $CE$. By angles chasing we have $$\angle PDE=\angle CPA', \angle DEP=\angle A'BC~\text{and}~\angle PEA'=\angle PBA'.$$ Applying Ceva's sine version on $\triangle EDA'$ and $\triangle BPC$, we obtain that $\angle PDA'=\angle PCA'$, implying $\angle TBA'=\angle CPA'=\angle CDA'$ , i.e. $\square TDA'B$ is cyclic. This implies $$\angle A'CT=\angle A'CD=\angle A'PB=\angle EBP=\angle TBD=\angle TA'D$$and hence $TC\cdot TD=TA'^2=TA^2$ since $TA=TA'$ by construction of $A'$. So we are done.

Let x be the intersection of bd and ce, and y be the intersection of bc and de. Note that abxe is a parallelogram (so ax is the a-median of abe) and yb, yc are the two tangents to the circumcircle of abe: therefore ay is the a-symmedian on abe: ax and ay are isogonal wrt bae. Now the isogonality lemma gives that ac and ad are also isogonal wrt bae: from here, law of sines in abc and ade yields ac:ad = bc^2:de^2. Menelaus on cdy and bet gives bc^2:de^2 = ct^2:dt^2. ac:ad=ct^2:dt^2 implies tad and tca are similar, as desired.

Let $S=BD\cap CE$, and let $U$ be the Miquel point of $TESD$. Then since $\measuredangle CBE=\measuredangle BED$, choose $C'$ such that $CBEC'$ is an isosceles trapezium. Then $$\measuredangle CC'D=\measuredangle BED=\measuredangle ESB=\measuredangle CSD$$so $CSDC'$ is cyclic. Since $U= (CSD)\cap (BSE)$, we know that the perpendicular bisector of $BE$ bisects $US$ so since $ABSE$ is a parallelogram, $U,A$ are symmetric in $BE$. Therefore $$\measuredangle TUD=\measuredangle TBD=\measuredangle EBD =\measuredangle UEB=\measuredangle USB=\measuredangle USD=\measuredangle UCD$$so by alternate segment theorem, $TU$ is tangent to $(UCD)$. Since $U,A$ are symmetric in $BE$ and $T\in BE$: $$TA^2=TU^2=TC\times TD$$so $TA$ is tangent to $(ACD)$.

this is harder than a G8 and i'm worried Let $BD$ and $CE$ intersect at $A'$. Note that $A'EAB$ is a parallelogram by conditions and the fact that its a convex pentagon. aaa. Let the line through $C$ parallel to $BE$ intersect line $ED$ at $C'$. Then $\providecommand{\dang}{\measuredangle} \dang DC'C = \dang EC'C = \dang C'EB = \dang DEB = \dang EAB = \dang BA'E = \dang DA'C$, whence $CC'A'D$ is cyclic and so the reflection of $A'$ over the perpendicular bisector of $BE$ lies on $(CA'D)$. Let this point be $A''$ and note that $A''$ and $A$ are reflections over $BE$. Now observe that $A''$ is the center of the spiral similarity sending $CD$ to $EB$, hence $A''$ lies on $(BDT)$. From here, just note that \[ \providecommand{\dang}{\measuredangle} \dang TA''D = \dang TBD = \dang EBD = \dang EBA' = \dang A''EB = \dang A''CD, \]so we find that $A''$ is tangent to $(A''CD)$. Therefore, we obtain $TA^2 = TA''^2 = TD \cdot TC$, as desired.

Let $BD \cap CE = A'$ and $BC \cap DE = L$. Since $\angle BCE = \angle BEA = \angle EBA'$ and $\angle BDE = \angle ABE = \angle BEA'$ we have $BE$ tangent to $(BEC)$ and analagously $(BED)$. Hence, we have $AE \cdot BD = BE^2 = AB \cdot EC$ or $\frac{AE}{EC}=\frac{AB}{BD}$. It follows that $\triangle AEC \sim \triangle ABD$. Then, by the law of sines we get $\frac{AC^2}{AD^2}=\frac{AE^2}{AB^2}=\frac{BC}{DE}$. Note that $BL=EL$ since $\angle EBL = \angle BEL$. Then, applying Menelaus' theorem on $\triangle CDL$ with respect to line $BET$ we get: $$\frac{DE}{EL} \cdot \frac{LB}{BC} \cdot \frac{CT}{TD}=1 \textrm{ and hence } \frac{AC^2}{AD^2}=\frac{BC}{DE}=\frac{TC}{TD}$$Now, suppose for contradiction that there is a point $X \neq A$ on line $TA$ such that $AXDC$ is cyclic. It is well known that $\frac{TC}{TD}=\frac{AC}{AD} \cdot \frac{XC}{XD}$ holds. It follows that $\frac{AC}{AD}=\frac{XC}{XD}$. Then, the inner angle bisectors of $\angle CAD$ and $\angle CXD$ must intersect on $CD$. But this cannot happen since they intersect on the midpoint of the arc $BC$ not containing $A$ of $(AXDC)$.

Revenge? Let $X=BC\cap ED$ be the intersection of the tangents to $(ABE)$ at $B$ and $E$ and let $a,b,e$ be the side lengths of $\triangle ABE$. Note $ED = \frac{ea}{b}, BD = \frac{a^2}{e}, BC = \frac{ba}{e}, EC = \frac{a^2}{b}$. In particular, $\triangle ABD\sim\triangle AEC$ by SAS. Now by Menelaus on $\triangle XCD$, $$\frac{TD}{TC} = \frac{BX}{BC}\cdot \frac{ED}{EX} = \frac{e^2}{b^2} = \left(\frac{AD}{AC}\right)^2$$ and we're done!

Solved with GrantStar. This problem is kinda broken – the statement with one condition deleted is susceptible to an affine transformation. Here is the modified statement: Quote: Let $ABCDE$ be a convex pentagon such that $AEDB$ and $ABCE$ are trapezoids with $AB \cdot CE = AE \cdot DB$. Lines $BE$ and $CD$ intersect at point $T$. Prove that line $AT$ is tangent to the circumcircle of $\triangle ACD$. (The original problem has more conditions because it reqiures $AB \cdot CE = AE \cdot DB = BE^2$.) Note that by SAS similarity, $\triangle ABD \sim \triangle AEC$, so the angle bisector of $\angle CAD$ is also the angle bisector of $\angle BAE$. Thus, vertically stretching the diagram about the bisector of $\angle BAE$ preserves the length condition in the (weaker) problem. It also preserves what we want to show, since the statement is equivalent to showing that the angle bisector of $\angle BAE$ is perpendicular to the angle bisector of $\overline{CD}$ and $\overline{AT}$. We do this vertical stretch until $\overline{BD} \perp \overline{AC}$, and due to the similarity we have $\overline{CE} \perp \overline{AD}$ as well: $X$, the intersection of $\overline{CE}$ and $\overline{DB}$ is now the orthocenter of $\triangle ACD$. Now, we can angle chase to see that $\triangle EAX \sim \triangle BXA \sim \triangle ACD$. It suffices to show that if $M$ is the midpoint of $\overline{AX}$ and $T$ is where the tangent to $(ACD)$ at $A$ meets line $BC$, then line $MT$ is perpendicular to the $A$-median in $\triangle ACD$. (This is probably well-known). This is a much more standard problem: letting $P$ be the $A$-humpty point in $\triangle ACD$, note that inversion at $T$ with radius $TA$ swaps $(APC)$ and $(APD)$, so $TP = TA$. From $AM=MH$, it follows that $M$ lies on the altitude from $T$ to $\overline{AP}$, proving the lemma.