Maybe just use the law of sines and calculate all the length of sides? We just need to prove that BI/BJ=DB*LI/(LB*ID).

Complex bash works very well too, I will give a sketch. Firstly, it's well-known that $L$ lies of $EF$ (this appears in ISL 2005 G6), so we can delete $M$. Now, let the incircle be the unit circle and let $d=1$. We have $a=\frac{2ef}{e+f}$ and $b=\frac{2f}{f+1}$ (we don't even need $c$). Since $A$ is midpoint of $IJ$, $j=2a-i=2a$. Since $L$ lies on $ID$ and $ID$ is the real axis, $l=\overline{l}$. Since $L$ lies on $EF$, $\frac{l-e} {l-f} = \overline {(\frac {l-e}{l-f})}$ and easy calculation gives $l=\frac {f+e} {fe+1}$. Now we shall check that $\frac{(i-b)(j-b)} {(l-b)(a-b)}=\overline{(\frac{(i-b)(j-b)} {(l-b)(a-b)})}$, which is a matter of direct computation.

Let the line that passes through $I$ and is perpendicular to $BI$ meets $AB$ and $BC$ at $X, Y$. Also let the line that passes through $I$ and is perpendicular to $CI$ meets $AC$ and $BC$ at $Z, W$. Since $\overline{IX}=\overline{IY}$, $\overline{IZ}=\overline{IW}$ $\rightarrow$ $\triangle{IXZ}\equiv \triangle{IYW}$. So $XZ$ is parallel to $BC$ and tangent to the incircle of $\triangle{ABC}$. Let the incircle meets $BC, CA, AB$ at $D, E, F$. Then by Brianchon's theorem at $BCEZXF$, $BE$, $CX$, $EF$ meet at a point $L'$. Since $\triangle{XL'Z} \sim \triangle{CL'B}$, $MT$ passes through $L'$ ($T$ is the midpoint of $XZ$). Because $MT$ passes through $A$, $L'$ is on $AM$ and $EF$ which meens that $L'=L$. (it is well known that $L$ is on $EF$). By easy angle chasing, $\triangle{BWI} \sim \triangle{BIA} \rightarrow \triangle{BWZ} \sim \triangle{BIL}$. So, $\angle{IBL}=\angle{IBZ}=\angle{ABJ}$.

Nice one Let perpendicular at $I$ to $IC$ meet $AC$ at $S$. Claim $: BAJ$ and $IBS$ are similar. Proof $:$ Note that $\angle BIS = \angle BIL + \angle LIS = \angle 90 + \frac{\angle B}{2} + \frac{\angle C}{2} = \angle 180 - \frac{\angle A}{2} = \angle BAJ$ and $\frac{IS}{AJ} = \frac{IS}{IA} = \frac{BI}{BA}$. So now we just need to prove $B,L,S$ are collinear. Note that $\frac{AL.BM}{LM.AB} = \frac{\sin{ABL}}{\sin{MBL}}$ and $\frac{AS.BC}{SC.AB} = \frac{\sin{ABS}}{\sin{CBS}}$. Since $\angle ABL + \angle MBL = \angle ABS + \angle CBS = \angle ABC < \angle 180$ then we only need to prove $\frac{AL.BM}{LM.AB} = \frac{AS.BC}{SC.AB}$ or $\frac{2AS}{SC} = \frac{AL}{LM}$ Let $T$ be where perpendicular bisector of $BC$ meets $AI$ and $I_a$ be excenter. Note that $\frac{AL}{LM} = \frac{AI}{IT} = \frac{2AI}{II_a}$ so we need to prove $\frac{2AS}{SC} = \frac{2AI}{II_a}$ or $\frac{AS}{SC} = \frac{AI}{II_a}$ which is true since $\angle ASI = \angle 90 + \frac{\angle C}{2} = \angle ACI_a$. we're Done.

As we all know $L \in EF$. In $\triangle BAI$ and $\triangle BAJ$ we have $$\frac{sin \angle ABJ}{sin \angle ABI}=\frac{AB \cdot sin \angle ABJ}{AB \cdot sin \angle ABI}=\frac{AJ \cdot sin \angle AJB}{AI \cdot sin \angle AIB}=\frac{sin (\frac{A}{2} - \angle ABJ )}{sin(\frac{\pi}{2} -\frac{C}{2})}$$So $$cot \angle ABJ=cot \frac{B}{2} +2cot \frac{A}{2}$$In $\triangle EIF$ we have $$IL= \frac{r \cdot sinA}{sinB+sinC}$$In $\triangle BIL$ we habe $$\frac{sin \angle LBI}{cos(\angle LBI + \frac{B}{2})}=\frac{sin \angle LBI}{sin \angle BLD}=\frac{IL}{IB}=\frac{IL}{\frac{r}{sin\frac{B}{2}}}$$So $$cot \angle LBI=cot \frac{B}{2} +2cot \frac{A}{2}=cot \angle ABJ . \square$$

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