The condition $BC = 3(AC-AB)$ implies $BD\cdot 2 = DC$. Moreover, since $E, F$ are reflections w.r.t. $\overline{AI}$, then $DE = KF$ and $\overline{DE} \cap \overline{KF} \in \overline{AI}$. Denote $M = \overline{AI} \cap \overline{BC}$, $N = \overline{KF} \cap \overline{DE} \cap \overline{AI}$, and $I_A$ as the $A$-excenter of $\triangle ABC$. Claim 01. $B, I, L, C$ are concyclic. Proof. \begin{align*} \measuredangle BLC &= \measuredangle BLK + \measuredangle KLC \\ &= \measuredangle BFK + \measuredangle KEC \\ &= \measuredangle DEC + \measuredangle BFD \\ &= \measuredangle DIC + \measuredangle BID = \measuredangle BIC. \end{align*}Claim 02. $\overline{IL} \perp \overline{LK}$. Proof. \begin{align*} \measuredangle ILK &= \measuredangle ILB + \measuredangle BLK \\ &= \measuredangle ICB + \measuredangle BFK \\ &= \measuredangle ICB + \measuredangle DEC = 90^{\circ}. \end{align*}Claim 03. $(N, I_A; M, I) = -1$. Proof. Let $G$ be the midpoint of $CD$. Therefore, $BD=DG=GC$, so the $A$-excircle is tangent to $\overline{BC}$ at $G$, i.e. $\overline{I_AG} \perp \overline{CD}$. This means $\triangle I_ACD$ is isosceles at $I_A$, so $\measuredangle I_ADC = \measuredangle DCI_A = \measuredangle CDE \implies DC$ is the angle bisector of $\measuredangle I_ADE$. Since $\overline{DI} \perp \overline{DM}$ and $\overline{DM}$ is bisector of $\angle I_ADN$, the conclusion follows. The second claim implies that $L, K, I_A$ are collinear. Moreover, if we let $K'$ as the second intersection of $\overline{KL}$ and the incircle of $\triangle ABC$, $L$ is the midpoint of $KK'$. Therefore, it suffices to prove $KF = KK'$. Since $\overline{KI} \perp \overline{KM}$ we only need to prove $(\overline{KF}, \overline{KL}; \overline{KM}, \overline{KI}) = -1$. However, projecting this pencil to line $\overline{AI}$ gives us \[ (\overline{KF}, \overline{KL}; \overline{KM}, \overline{KI}) = (N, I_A; M, I) = -1, \]as desired. $\square$

Seicchi28 wrote: The condition $BC = 3(AC-AB)$ implies $BD\cdot 2 = DC$. Moreover, since $E, F$ are reflections w.r.t. $\overline{AI}$, then $DE = KF$ and $\overline{DE} \cap \overline{KF} \in \overline{AI}$. Denote $M = \overline{AI} \cap \overline{BC}$, $N = \overline{KF} \cap \overline{DE} \cap \overline{AI}$, and $I_A$ as the $A$-excenter of $\triangle ABC$. Claim 01. $B, I, L, C$ are concyclic. Proof. \begin{align*} \measuredangle BLC &= \measuredangle BLK + \measuredangle KLC \\ &= \measuredangle BFK + \measuredangle KEC \\ &= \measuredangle DEC + \measuredangle BFD \\ &= \measuredangle DIC + \measuredangle BID = \measuredangle BIC. \end{align*}Claim 02. $\overline{IL} \perp \overline{LK}$. Proof. \begin{align*} \measuredangle ILK &= \measuredangle ILB + \measuredangle BLK \\ &= \measuredangle ICB + \measuredangle BFK \\ &= \measuredangle ICB + \measuredangle DEC = 90^{\circ}. \end{align*} I was missing Seicchi's claims 1 and 2 in my original attempt, so I will quote them here(and use them). We instead define $N$ to be the midpoint of $DE$. Observe that if $DE$ = $2LK$, we have $ID = IK$, $DN = KL$ and $\angle ILK = \angle IND = 90^{\circ}$, and hence $\triangle ILK \cong \triangle IND \sim IEC$. Hence, we wish to prove that $\triangle ILK \sim \triangle IEC$. At most 2 choices of $L$ satisfies $\angle ILK = 90^{\circ}$ and $L, K, E, C$ concyclic, as 2 different circles intersect at at most 2 points. One of them is $K$. We first show that $L \neq K$. Let the reflection of $C$ across $AI$ be $C'$. Clearly $BC' = BD = BF$ so $\angle IKC = \angle IDC' = 90^{\circ} + \angle FDI = 90^{\circ} + \frac{1}{2}\angle B$, implying $B,I,K,C$ isn't convex and hence isn't concyclic so by claim 1 $L \neq K$. Next we show that the choice $L'$ such that $\triangle IL'K \sim \triangle IEC$ satisfies $L', K, E, C$ concyclic, or that $\angle LEK + \angle LKC = 180^{\circ}$, which will imply that $L = L'$. We have the following: $$\angle IDC' = 90^{\circ} + \angle FDI = 90^{\circ} + \frac{1}{2} \angle B$$$$\angle C'ID = \angle CID - \angle MIK = \frac{1}{2}\angle A + \frac{1}{2}\angle C - (\frac{1}{2}\angle B - \frac{1}{2}\angle C)$$$$\angle IC'D = 180^{\circ} - \angle C'ID - \angle IDC' = (\frac{1}{2}\angle B - \frac{1}{2}\angle C)$$$$ \angle L'KC = \angle IKC - \angle IKL' = \angle IDC' - \frac{1}{2}\angle C = 90^{\circ} + (\frac{1}{2}\angle B - \frac{1}{2}\angle C)$$\begin{align*} \angle L'EC &= 90^{\circ} - \angle IEL' = 90^{\circ} - \angle ICK \\ &= 90^{\circ} - \angle ICD' = 90^{\circ} - (\frac{1}{2}\angle B - \frac{1}{2}\angle C) \end{align*}where the last used spiral similarity on $I, L', E, K, C$. $\angle L'EC + \angle L'KC = 180^{\circ}$ and hence $L = L'$ and we're done.

This one was proposed by me (wow, I realised quite late that BMO SL is available on AoPS) Anyway, this is the solution I sent together with the problem: Define by $f$ the reflection over the line $\overline{AI}$. Let $f(B)=B'$ and $f(C)=C'$, so $B'\in \overline{AC}$ and $C'\in \overline{AB}$. We also know that $f(D)=K$. Then $f$ sends the circle $(BFK)$ to circle $(B'ED)$ and the circle $(CEK)$ to circle $(C'FD)$. If circles $(B'ED)$ and $(C'FD)$ cut each other again at $T$, then $f(L)=T$. Moreover, $K$ lies on the incircle of $ABC$, since $D$ lies too and $\overline{AI}$ passes through the centre of the incircle. Now we shall use the condition $\frac 13 BC=AC-AB$. It rewrites as $2(BC+AB-AC)=BC+AC-AB$, which translates into $2BD=CD$, since $BD=p-b$ and $CD=p-c$. $f(F)=E,f(B)=B'$, so $BF=B'E$ or $B'E=BF=BD=\frac 12 CD=\frac 12 CE$, thus $B'$ is the midpoint of line segment $CE$. Since $\overline{BD}$ and $\overline{BE}$ are tangent to the incircle of $ABC$, it follows that $\overline{B'K}$ and $\overline{B'E}$ are tangent to the incircle of $ABC$, so $B'K=B'E=B'C$, which leads to $\angle CKE=90^o$. If $Q$ is the midpoint of line segment $DE$, we have $\angle CQE=90^o$, so $C,Q,K,L,E$ are concyclic. Next, we prove that $L$ lies on the circle $(BIC)$. $\angle BLC=\angle BLK+\angle CLK=\angle BFK+\angle CEK=360^o-\angle AFK-\angle AEK=\angle BAC+\angle FKE=\angle BAC+\angle FDE=\angle BAC+90^o-\frac 12 \angle BAC=90^o+\frac 12 \angle BAC=\angle BIC$, so $B,I,L,C$ are concyclic, which means $T$ lies on $(BIC)$ too, since $\overline{AI}$ passes through its centre, the midpoint of arc $BC$ in $(ABC)$. Our next claim is that $\angle ILK=90^o$. Indeed, it is equivalent to proving that $\overline{KL}$ passes through $J$, the $A-$excentre of $ABC$, since $J$ is the antipode of $I$ in $(BIC)$. To prove this, just note that $\angle CLJ=\angle CBJ=90^o-\frac 12 \angle ABC=\angle BFD=\angle CEK=\angle CLK$, so $K,L$ and $J$ are collinear. Since $B'$ is the centre of $(CEK)$, $B$ is the centre of $(FDC')$. $\angle ILK=90^o$ implies $\angle ITD=90^o$, so if $S\in \overline{IT}\cap \overline{BC}$, then $\angle DTS=90^o$ which leads to $S$ being on the circle $(FDC')$. Then $BS=BD$, so $CD=2BD$ implies $CD=DS$. It follows that triangles $\Delta IDC\equiv \Delta IDS$, which implies that the heights $DQ=DT$. Comsequently, $DE=2DQ=2DT=2KL$, as required.

Attachments:

Let $B_1$ and $C_1$ be reflections of $B$ and $C$ across $AI$. Claim $: CD = 2BD$. Proof $:$ The Whole point is to translate the $\frac{1}{3} BC = AC - AB$ into something useful. $\frac{1}{3} BC = AC - AB \implies a + c - b = \frac{a + b - c}{2} \implies 2(p-b) = p-c \implies CD = 2BD$ Claim $: BILC$ is cyclic. Proof $:$ $\angle BLC = \angle BLK + \angle CLK = \angle BFK + \angle CEK = \angle B_1ED + \angle C_1FD = \angle BFD + \angle CFD = \angle BID + \angle CID = \angle BIC$. Claim $: L,K,I_a$ are collinear. Proof $:$ Note that $BILCI_a$ is cyclic so $\angle BLI_a = \angle BII_a = \angle CID = \angle CED = \angle B_1ED = \angle BFK = \angle BLK$ Let $I_aD$ meet $BIC$ at $S$. Let $CI$ meet $DE$ at $R$. Let $T$ be point such that $I_aT \perp BC$. Since $KL = DS$ and $DR = \frac{DE}{2}$ then we need to prove $DR = DS$. Let $C'$ be reflection of $C$ across $D$. Note that since $\angle ISD = \angle ISI_a = \angle 90 = \angle DRI$ so we just need to prove $I,S,C'$ are collinear or in fact we need to prove $\angle DSC' = \angle 90$. Claim $: C'STI_a$ is cyclic. Proof $:$ Note that $BD = CT \implies DT = CT \implies C'D.DT = CD.BT = DS.DI_a \implies C'STI_a$ is cyclic. Now we have $\angle DSC' = \angle I_aSC' = I_aTS = \angle 90$. we're Done.

The length condition is equivalent to $CD = 2 \cdot BD$. Let the tangent to the incircle at $K$ intersect $AC, AB$ at $X, Y$ respectively; then $B$ and $X$ are the midpoints of segments $FY$ and $EC$. Furthermore, if $W$ is the midpoint of $KY$ then $KWBF$ is a cyclic isosceles trapezoid, and if $Z$ is the reflection of $K$ over $X$ then $EKCZ$ is a rectangle. So, $XL = XK = YB = YW \implies L$ is the reflection of $B$ over the perpendicular bisector of $WK$, so $KL = BW = \frac{FK}{2} = \frac{DE}{2}$. $\square$

oops this took a while (last part was embarrassingly long); once again intuition is bad (need to stop avoiding things subconciously!) else it wouldve been like a 20 minute solve Note that $CD=BD\cdot 2$. Reflect $B,C,K$ across $AI$ to $B',C',D$, and also define a point $C''$ on ray $AB$ past $B$ such that $BC=BC''$; in particular, $FC''$ has midpoint $C'$. Let $U$ be the midpoint of $DE$, and let $L'$ be the reflection of $U$ across $DI$. Let $F'$ be the reflection of $F$ across $DI$. I claim that $L'$ is the reflection of $L$ across $AI$; it suffices to show that $L'=(B'ED)\cap (C'FD)$. Since $B'$ is the midpoint of $CE$ from lengths, we find that $B'UL'$ are collinear, and we also find that $B'E=\frac{1}{2}CE=\frac{1}{2}CD=B'U$ implying that $\triangle B'EU$ and $\triangle DL'U$ are isosceles, so $B'EL'D$ is cyclic. Now I claim that $F'EUD\sim DFC'C''$. Note that: \[\angle DFC''=\angle DEF=\angle DEF'\]and \[\angle FDC''=\angle FDB+\angle BC'C=\angle DEF+\angle AFE=\angle DEF+\angle FDE=\angle EF'D\]so the claim is proven. But now: \[\angle FC'D=\angle EUF'=180^{\circ}-\angle DUF'=180^{\circ}-\angle DL'F\]so $L'\in (C'FD)$. So $KL=DL'=\frac{1}{2}DE$. Done! (not using directed angles because I'm very lazy)