Wow, I hope this is okay. Let $O$ be the midpoint of $BC$. Let's animate $D$ over the (fixed) perpendicular bisector of $AB$. Obviously $OD, OE$ are both perpendicular bisector so $\angle{DAO} = \angle{OBD} = \angle{CBD} = \angle{ECB} = \angle{OAE}$ which means that in fact $E$ is just the result of 2 processes: 1. First reflect line $AD$ over fixed line $AO$. 2. This new line, project it (perspective from $A$) over fixed perpendicular bisector of $AC$. So we get a projective map $D \mapsto E$. By definition, $\deg{D} = 1 = \deg{E}$. But (Moving Points Lemma) says that line $DE$ has degree at most $\deg{D} + \deg{E} - \text{\# coincidences}$. In case $D = O$ then we also get $E = O$ so $\text{\# coincidences} \geq 1$ implying that line $DE$ has degree at most $1$. So to finish the problem we just need at least 2 different configurations. $\,\,\,\, \star \,\, DB \perp BC$, in this case we also have $EC \perp BC$ which means that $D, A, E$ are collinear and tangent to $(ABC)$... the conclusion follows. $\,\,\,\, \star \,\, D = \infty$, in this case we have $E$ being on the line $CA'$ (also by definition $OE \perp AC$) where $A' := $ reflection of $A$ over $BC$. We want to prove that $ET \perp AB$. To complete this config, is just angle chasing. Let $\theta := \angle{ACO}$. We have $\angle{EAO} = \angle{OAC} = \theta = \angle{ACO} = \angle{OCE}$. So we're done if we prove that $T, A, O, E$ are concyclic, which in turn means proving $\angle{OTA} = \angle{OEA}$. And that's it because we already know $\angle{TAC} = \angle{AOE}$ (look at triangles $\triangle{TAC}, \triangle{AOE}$). We're done! P. S. In fact last case is analogous if $E = \infty$ so we even have proved the result for 3 different configurations.

Let $\angle DBC=\angle ECB=x$. Note that $\angle TBD=180^\circ-x$ and $\angle TAD=\angle TAB+\angle BAD=\angle C+\angle ABD=\angle B+\angle C-x=90^\circ-x$ and $\angle TCE=x$ and $TAE=\angle TAC-\angle EAC=90^\circ+\angle C-\angle ECA=90^\circ+x,$ and so by applying LoS in triangles $TBD,TAD$ and $TEC,TEA$, we may infer that $\dfrac{\sin \angle DTB}{\sin \angle DTA}=\dfrac{\sin x}{\sin(90^\circ-x)}=\dfrac{\sin x}{\sin(90^\circ+x)}=\dfrac{\sin \angle ETC}{\sin \angle ETA},$ hence if we denote $\angle DTB=a$, $\angle ETC=b$ and $\angle ATB=s$, we obtain $\dfrac{\sin a}{\sin(s-a)}=\dfrac{\sin b}{\sin(s-b)},$ which, due to the monotonicity of the function $\dfrac{\sin x}{\sin(s-x)}$ implies that $a=b$, i.e. $T,D,E$ are collinear, as desired.

Lukaluce wrote: Let $ABC$ be a right triangle with hypotenuse $BC$. The tangent to the circumcircle of triangle $ABC$ at $A$ intersects the line $BC$ at $T$. The points $D$ and $E$ are chosen so that $AD = BD, AE = CE,$ and $\angle CBD = \angle BCE < 90^{\circ}$. Prove that $D, E,$ and $T$ are collinear. Proposed by Nikola Velov, Macedonia A great problem!!! Obviously, there is a circle $w$ that touches $BD,CD$ at points $B,D$. $Pow(T,w) = TB*TC = TA^2 = Pow(T, \odot (A))$ and $Pow(D,w) = BD^2 = AD^2 = Pow(D,\odot(A))$ and $Pow(E,w) = CE^2 = AE^2 = Pow(E,\odot(A)) => D,E,T$ collinear.

Attachments:

We have that $(AD, AO, AE, AT)$ is harmonic and $(AB, AH, AC, AT)$ is also harmonic, where $H$ is the foot of the perpendicular from $A$ to $BC$. The perpendiculars from $O$ to the lines of $(AB, AH, AC, AT)$ also form an harmonic bundle, therefore $(OD, OA, OE, OT)$ is harmonic. Since $(AD, AO, AE, AT)$ and $(OD, OA, OE, OT)$ are harmonic, their intersection form an harmonic quadruple, therefore $D, E, T$ collinear, as desired.

Amusingly, this was also independently discovered by Malay and Sidharth and appeared as LMAO P2 this year.

L567 wrote: Amusingly, this was also independently discovered by Malay and Sidharth and appeared as LMAO P2 this year. Can I find out what kind of competition it is?

Let $M$ be the midpoint of $BC$, $X := DM\cap AT$ and $Y := EM\cap AT$. Then $\triangle XMY\sim\triangle BAC$, and the points $T,X,A,Y$ are harmonic in that order, that is, $\frac{XT}{TY}=\frac{XA}{AY}$. Thus, to prove that $T, D$ and $E$ are collinear, it suffices to show that $XE, DY$ and $AM$ concur (as the desired conclusion follows from Menelaus' theorem). Let $x=\angle C$ and $\alpha=\angle CBD=\angle BCE$. Note that quadrilaterals $XBMA$ and $YCMA$ are cyclic. By angle chasing and sine law, we have \[\begin{aligned} \dfrac{MD}{DX}&=\dfrac{MD}{BD}\cdot\dfrac{BD}{DX}=\dfrac{\sin\alpha}{\sin x}\cdot\dfrac{\sin(90^\circ-x)}{\sin(90^\circ-\alpha)} = \tan\alpha\cot x,\\ \dfrac{ME}{EY}&=\dfrac{ME}{CE}\cdot\dfrac{CE}{EY}=\dfrac{\sin\alpha}{\sin(90^\circ-x)}\cdot\dfrac{\sin x}{\sin(90^\circ-\alpha)} = \tan\alpha\tan x. \end{aligned}\]Since $\frac{XA}{AY}=\frac{XM^2}{MY^2} = \frac{AB^2}{AC^2} = \tan^2 x$, we get \[\frac{EY}{ME}\cdot\frac{MD}{DX}\cdot\frac{XA}{AY} = \cot\alpha\cot x\cdot\tan\alpha\cot x\cdot\tan^2 x=1.\]Hence, by the converse of Ceva's theorem, $XE, DY$ and $AM$ concur as desired.

Too easy. Let $R$ be the intersection of $BD$ and $EC$ and let $M$ be the midpoint of $BC$, then $RM$ is perp. to $BC$, $MD$ perp. to $AB$ and $ME$ is perp to $AC$. Let's prove the converse of Menelaus theorem according to points $T$, $D$, $E$ in triangle $RBC$. We can find by applying ratio lemma in triangles $BRM$ and $MRC$, that: $(CE/ER)×(RD/DB)=(sin(\angle B)/sin(\angle C)^2$, and we can easily see that $BT/CT=(BT×CT)/(CT^2)=(AT/CT)^2=(sinC/sinB)^2$, which implying that: $(TB/TC)×(CE/ER)×(RD/DB)=1$, which completing the proof.

Let $M$ be midpoint of $BC$. Perform inversion at center $M$ with radius $MB$.Let inverse of point $Z$ is $Z'$. Since $\angle MAT=90$,we have $T'$ is feet of perpendicular from $A$to $BC$. Let $CE'$ meets $AB$ at $X$ and $BD'$ meets $CA$ at $Y$.We have, $\angle BXC=\angle ME'C=\angle MEC=\angle MBD=\angle MD'B=\angle BYC$ $\implies$ $BYXC$ cyclic. Since $D,E,T$ collinear iff and only if $D'MT'E'$ is cyclic. But this is wellknown fact, since we have $BYXC$ cyclic quadirateral,dioganals are perpendicular and $M,E',T'$ are midpoints of its three sides.$\implies$ $T'$ lies on $(ME'T')$ Which is completes the proof.

Here is a very amazing and instructive projective solution presented by Rijul Saini sir. There are two parts to this solution, the first one is some patchy angle chasing for some important setup and the second part is the interesting projective part which I admire the most and its a very genius step! Define $O$ to be the midpoint of $BC$. Claim. $OA$ bisects $\angle DAE$. Proof. Note that \[ \measuredangle DAE = 180-\measuredangle DAB-\measuredangle EAC = 180-\measuredangle DBA -\measuredangle ECA = 2\measuredangle DBC\]by the given angle condition, where we use directed angles to avoid configuration issues. Therefore, as triangles $OAE$ and $OCE$ are congruent, we see that $\angle OAE = \angle OCE$ so we are done. Moving ahead to the more interesting part, we see what $OA$ and $TA$ are the internal and external angle bisectors respectively, of $\angle DAE$, so we must have that the pencil \[(AD,AE;AT,AO)\]is a harmonic bundle. Project this pencil on $DE$, and assume that the projection of $AT$ onto $DE$ is some point $T_0$, and that of $AO$ is $T'$, then, we see that \[(D,E;T',T_0)=-1\]Project this bundle on the $A$-midline, then we see that if $M$,$N$ are the midpoints of $AB,AC$, then, \[-1 = (M,N;OA\cap MN,\infty_{MN}) = (D,E;T',T_0)\]Therefore, we see that $AT_0\parallel BC$ as well as $T_0\in AT$, so $T_0\equiv T$ and we are done.

Let $BD\cap CE=F$ We can redefine $D=BF\cap MN,E=CF\cap MP$ where $N,P$ are midpoints of $AB,AC$ respectively. We will use moving points method. Animate $F$ over the perpendicular bisector of $BC$. Define the projective transformation as \[f: F\rightarrow BF\rightarrow BF\cap MN=D\rightarrow TD\rightarrow TD\cap FM\]\[g:F\rightarrow CF\rightarrow CF\cap MP=E \rightarrow TE\rightarrow TE\cap FM\]Where $FM$ is a constant line. $f,g$ has degree $2$ thus we will show that $f,g$ are same at $3$ different cases. $i)F=M$ $f:M\rightarrow BM\rightarrow BM\cap MN=M\rightarrow TM\rightarrow TM\cap MF=M$ $g: M\rightarrow CM\rightarrow CM\cap MP=M\rightarrow TM\rightarrow TM\cap MF=M$ Which are same. $ii)F=K$ where $K=AB\cap FM$ $f:K\rightarrow BK\rightarrow N\rightarrow TN\rightarrow TN\cap FM$ $g:K\rightarrow CK\rightarrow CK\cap MP=S\rightarrow TS\rightarrow TS\cap FM$ \[\frac{TB}{TC}.\frac{CS}{SK}.\frac{KN}{NB}=\frac{AB^2}{AC^2}.\frac{\frac{AC}{AB}}{\frac{AB}{AC}}=1\]Thus $f,g$ are same when $F=K$. $iii)F=L$ where $L=AC\cap FM$ $f:L\rightarrow BL\rightarrow BL\cap MN=X\rightarrow TX\rightarrow TX\cap MF$ $g:L\rightarrow CL\rightarrow P\rightarrow TP\rightarrow TP\cap MF$ \[\frac{TB}{TC}.\frac{CP}{PL}.\frac{LX}{XB}=\frac{AB^2}{AC^2}.\frac{AC^2}{AB^2}=1\]Thus $f,g$ are same again when $F=L$ as desired.$\blacksquare$

Let $\omega_1 = (D, DB), \omega_2 = (E, EC).$ $B \neq F = \omega_1 \cap \overline{BC}, G \neq C = \omega_2 \cap \overline{BC}$. $\theta := \angle{DBC} = \angle{BCE}$. Claim 1. $TA$ is tangent to $(AFG)$. Proof. $$\angle{AFG} = \angle{AFB} = \frac{\angle{ADB}}{2} = 90^{\circ} - \angle{ABD} = 90^{\circ} - \angle{B} + \theta$$and $$\angle{TAG} = \angle{TAC} - \angle{GAC} = \angle{C} + \angle{A} - (90^{\circ} - \angle{ECG}) = 180^{\circ} - \angle{B} - 90^{\circ} + \theta = 90^{\circ} - \angle{B} + \theta.$$Thus, $\angle{AFG} = \angle{TAG}$ and we are done with this claim. $\ \blacksquare$ Claim 2. $T$ is similitude center of $\omega_1$ and $\omega_2$. Proof. Consider the inversion from $T$ with radius $TA$. Due to the first claim, we have $TA^2 = TG \cdot TF$. Therefore, $G$ and $F$ will swap the places. Clearly, $B$ and $C$ will also swap the places and $A$ will stay the same. Thus, $\omega_1 = (ABF)$ and $\omega_2 = (ACG)$ will swap the places. It is well-known that if there exists an inversion that swaps two circles, then center of the inversion is also one of the centers of similitude of those two circles and thus we are done. $\ \blacksquare$ Claim 2 immediately implies that $T, D, E$ are collinear. $\square$ Remark. Triangle being right was not necessary.