First of all we note that $T$ is the first Fermat Point of $\triangle ABC$ It is known that in a triangle with $\angle BAC=60^\circ$, then the first Fermat Point is the $A-Dumpty Point$ (Mid point of the Symmedian Chord) Due to this $\angle TAC=\angle TBA=x $ and $\angle TAB = \angle TCA=y$ Let $G$ and $H$ be perpendiculars from $K$ and $L$ onto $AT$ To Prove that $KG=LH \iff AK \cdot \sin x=AL \cdot \sin y$ $$\iff \dfrac{AK}{AL}=\dfrac{\sin y}{\sin x} \text{, but } AK \cdot AC=AB \cdot AL \text{ (due to Power of Point)}\implies \dfrac{AK}{AL}= \dfrac{AB}{AC}$$Hence it suffice to show that $\dfrac{AB}{AC}=\dfrac{\sin y}{\sin x}$ But this follows directly from the Isogonal Ratio Lemma and Sine rule, since $AT$ is symmedian.

This was so cutee uwu Sharygin P11 wrote: Let $ABC$ be a triangle with $\angle A=60^o$ and $T$ be a point such that $\angle ATB=\angle BTC=\angle ATC$. A circle passing through $B,C$ and $T$ meets $AB$ and $AC$ for the second time at points $K$ and $L$.Prove that the distances from $K$ and $L$ to $AT$ are equal. Note that $\angle ATB=\angle CTA=120\implies \angle CAT=60-\angle TAB=\angle TBA\implies \Delta ATB\sim\Delta CTA.$ Hence $$\frac{AT}{TB}=\frac{CT}{AT}.$$Now, since $BCTLK$ is cyclic. So $\angle BLC=180-\angle BTC=60\implies ALB\text{ is equilateral. }\implies AL=AB.$ Similarly $AK=AC.$ Define $G,N,M,I$ as the feet of altitude to $AT$ from $K,C,B,L$ respectively. Note that $$\frac{KG}{BM}=\frac{CN}{LI}=\frac{AC}{AB}$$Note that $$\frac{KG}{BM}\times\frac{CN}{LI}=\left(\frac{AC}{AB}\right)^2$$Lemma: $$\frac{CN}{BM}=\left(\frac{AC}{AB}\right)^2$$ Proof: Note that $\Delta CTN\sim \Delta BTM $ as $\angle CTN= \angle BTM, \angle CNT=\angle BMT=90\implies$ $$\frac{CN}{BM}=\frac{BT}{CT}$$Note that $$\frac{AB}{AC}=\frac{BT}{AT}=\frac{CT}{AT}\implies \frac{BT}{CT}=\left(\frac{AC}{AB}\right)^2$$But $\frac{CN}{BM}=\frac{BT}{CT}.$ Hence $$\frac{CN}{BM}=\left(\frac{AC}{AB}\right)^2$$ Now, since $$\frac{CN}{BM}=\left(\frac{AC}{AB}\right)^2,\frac{KG}{BM}\times\frac{CN}{LI}=\left(\frac{AC}{AB}\right)^2\implies \frac{KG}{LI}=1.$$

It take me too long to realise that $T$ is nothing but Dumpty point , after this it was a 2 line sine rule bashh.. Here's my solution from actual contest...

Wait I believe you don't even need the Dumpty point. Let $AT$ meet $(BTC)$ for the second time at $T'$. Easy angle chase shows $AKT'L$ is a parallelogram, so $ALT'$ and $AKT'$ have equal areas, done.

toricelli and symmedians are nice

Interesting Notice that $T$ is the A-Dumpty point of $\triangle ABC$ We need to prove that $AK \cdot \sin \angle TAK = AL \cdot \sin \angle BAT$ and we know that $AK \cdot AC = AL \cdot AB$ which means we have to prove $$\frac{\sin \angle CAT}{\sin \angle BAT} = \frac{AL}{AK} = \frac{AC}{AB}$$which is immediate because the $A$-Dumpty point lies on the $A$-Symmedian $\blacksquare$

Let $A_1$ be the point such that $A, A_1$ are on opposite sides of $BC$ and $\triangle A_1BC$ is equilateral. Let $\omega$ be the circumcircle of $\odot (A_1BC)$. Then $T = A_1A \cap \omega \not \equiv A_1$. Furthermore, let $P, Q$ be the feet of the perpendiculars from $K, L$ to line $AT$. Lemma: $ALA_1K$ is a parallelogram. Proof of lemma: It suffices to show $AK \parallel LA_1$ and $AL \parallel KA_1$. We have $\angle A_1LC = \angle A_1BC = 60^{\circ} = \angle BAC$, so $AK \parallel LA_1$. The other case is analogous. $\blacksquare$. Since $ALA_1K$ is a parallelogram, we have $AL=A_1K$ and $\angle QAL = \angle PA_1K$. Since $\angle AQL = \angle A_1PK = 90^{\circ}$, we conclude that $\triangle AQL \cong \triangle A_1PK$, which gives $LQ=LP$, as desired.

Note that $T$ is the Fermat point of $\Delta ABC$. Let $P$ be the first isodynamic point of $\Delta ABC$. Recall that $T,P$ are isogonal conjugates. $$d(K,AT)=d(L,AT) \iff AK \sin \angle KAT = AL \sin \angle LAT \iff AC \sin \angle BAT = AB \sin \angle CAT \iff \frac{AB}{AC}=\frac{\sin \angle CAP}{\sin \angle BAP}$$Recall that the trilinear coordinates of $P$ are $(\sin (\angle A+\frac{\pi}{3}), \sin (\angle B+\frac{\pi}{3}), \sin (\angle C+\frac{\pi}{3}))$, so we have $$\frac{\sin \angle CAP}{\sin \angle BAP}=\frac{\sin (\angle B+\frac{\pi}{3})}{\sin (\angle C+\frac{\pi}{3})}=\frac{\sin \angle C}{\sin \angle B}=\frac{AB}{AC}$$so we are done.