CGMO, 2022

Proof. We need the following lemma.
Lemma: In triangle ABC, let I be the incenter. The incircle $\odot I$ is tangent to sides $BC$, $CA$, $AB$ at points $A_1$, $B_1$, $C_1$, respectively. Let $A_1I$ intersect $B_1C_1$ at $L$. Then, $AL$ passes through the midpoint $M$ of $BC$.

Proof of lemma: Draw a line through $L$ parallel to $BC$, intersecting $AC$ at $X$ and $AB$ at $Y$. Connect $IB_1$, $IC_1$, $IX$, and $IY$.
Since $A_1L \perp BC$ and $XY \parallel BC$, we know that $A_1L \perp XY$. Thus, $\angle ILX = \angle IB_1X = 90^\circ$, and $I, L, X, B_1$ are concyclic. Similarly, $I, L, C_1, Y$ are concyclic. Therefore, $\angle IXB_1 = \angle ILB_1 = \angle IYC_1$.
Combining this with $\angle IB_1X = \angle IC_1Y = 90^\circ$ and $IB_1 = IC_1$, we have $\triangle IB_1X \cong \triangle IC_1Y$. Thus, $IX = IY$.
Also, note that $IL \perp XY$, so $L$ is the midpoint of $XY$. Combined with $XY \parallel BC$, we conclude that $AL$ passes through the midpoint $M$ of $BC$. The lemma is proven.

Returning to the original problem, let the incircle $\odot I$ of triangle $ABC$ be tangent to sides $BC$, $CA$, $AB$ at points $A_1$, $B_1$, $C_1$, respectively. By the lemma, $L$ is on segment $B_1C_1$. Let $S$ be the midpoint of $BI$, and let $T$ be a point on $AS$ such that $\angle TBS = \angle BAS$.

We will now prove that points $B$, $T$, $L$ are collinear.

Since $\angle TBS = \angle BAS$, we know $\triangle BST \sim \triangle ASB$. Thus, $BS^2 = SA \cdot ST$. Also, $BS = SI$, so $IS^2 = SA \cdot ST$. This implies $\triangle IST \sim \triangle ASI$, and therefore,
$$\angle BTI = \angle BTS + \angle ITS = \angle ABI + \angle AIB = 180^\circ - \angle BAI.$$
Let $H$ be the orthocenter of triangle $AIB$. By the properties of the orthocenter, $\angle BHI = \angle BAI$. Thus, points $B$, $T$, $I$, $H$ are concyclic. Therefore, $\angle STH = \angle BTH - \angle BTS = \angle BIH - \angle ABI = 90^\circ$, and combined with $\angle AC_1H = 90^\circ$, we have points $A$, $C_1$, $T$, $H$ concyclic, with $AH$ being the diameter of the circle.
Thus,
$$\begin{aligned} \angle ITC_1 &= \angle HTC_1 - \angle HTI = 180^\circ - \angle HAC_1 - \angle HBI \\ &= 180^\circ - (90^\circ - \angle ABI) - (90^\circ - \angle ABI - \angle BAI) \\ &= 2\angle ABI + \angle BAI. \end{aligned}$$
Moreover, note that
$$\angle ILB_1 = \angle IC_1L + \angle LIC_1 = \frac{1}{2}\angle BAC + \angle ABA_1 = \angle BAI + 2\angle ABI.$$
So $\angle ITC_1 = \angle ILB_1$, so $I$, $T$, $C_1$, $L$ are concyclic. From this, we deduce $\angle ITL = \angle IC_1L = \angle BAI$. Therefore, points $B$, $T$, $L$ are collinear, and $\angle LBI = \angle TBI = \angle BAS$.
Since $AS$ is the median of triangle $BIJ$, we have $AS \parallel BJ$, and $\angle BAS = \angle ABJ$. Therefore, $\angle ABJ = \angle LBI$. The conclusion is proven. $\square$