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The incircle of a triangle $ABC$ has center $I$ and touches $AB , BC , CA$ at $C_1 , A_1 , B_1$ respectively. Let $BI$ intersects $AC$ at $L$ and let $B_1I$ intersects $A_1C_1$ at K. Show that $KL$ is parallel to $BB_1$.

I’ve draw some lines and try to find angles but it didn’t work. Can anyone give me some hints (or solution) please. Thank you very much!

(I didn’t try trigonometry or complex bash yet because it would be very ugly.)

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Here is my answer.

Since $AI$ and $CI$ are bisectors of $BAL$ and $BCL$ so we have $$\dfrac{IL}{IB}=\dfrac{AL}{AB}=\dfrac{CL}{CB}=\dfrac{AC}{AB+BC}$$

Now $\dfrac{IK}{IB_1}=\dfrac{S_{IA_1 C_1}}{S_{IA_1 B_1}+S_{IB_1 C_1}}$. We have $S_{IA_1 B_1}=\frac{1}{2}IA_1 IB_1 sinB_1IA_1= \frac{1}{2}IA_1 IB_1 sinC$. So $\dfrac{IK}{IB_1}=\dfrac{sinB}{sinA+sinC}$

We know that $\frac{AB}{sinC}=\frac{BC}{sinA}=\frac{CA}{sinB}$ so $\frac{IL}{IB}=\frac{IK}{IB_1}$, which gives the desired result.

Maybe there is another way without using Sin cos blah blah but I do not know. Hope this answer is helpful.

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Let $BL$ intersect $A_1C_1$ in point $M$. Then $M$ is the midpoint of $A_1C_1$ and $BL$ is the orthogonal bisector of $A_1C_1$. In the right angled triangle $IA_1B$, we have that $$IM \cdot IB = IA_1^2$$ But $IA_1 = IB_1 = IC_1$ so $$IM \cdot IB = IB_1^2$$ which can be reformulated as $$\frac{IM}{IB_1} = \frac{IB_1}{IB}$$ which means that triangles $IB_1M$ and $IBB_1$ are similar and therefore $$\angle \, IB_1M = \angle\, IBB_1$$ Furthermore, $IM \, \perp \, A_1C_1$ and $IB_1 \, \perp \, AC$ so $\angle\, LB_1K = 90^{\circ} = \angle \, LMK$. Consequently, the quadrilateral $LB_1MK$ is inscribed in a circle, and thus $$\angle \, KLB = \angle \, KLM = \angle \, KB_1M = \angle \, IB_1M= \angle \, IBB_1 = \angle \, LBB_1$$ which means that the lines $KL$ and $BB_1$ are parallel.

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