Quote: Let $ABCDEF$ be a bicentric hexagon. Let $\omega_{A}$ be the incircle of triangle $FAB$; define $\omega_B$, $\omega_C$, $\omega_D$, $\omega_E$, $\omega_F$ similarly. Let $\ell_{AB}$ be the external common tangent of $\omega_{A}$ and $\omega_{B}$ different from line $AB$; define $\ell_{BC}$, $\ell_{CD}$, $\ell_{DE}$, $\ell_{EF}$, $\ell_{FA}$ similarly. Let $A_1$ be the intersection point of the lines $\ell_{FA}$ and $\ell_{AB}$; define $B_1$, $C_1$, $D_1$, $E_1$, $F_1$ similarly. Prove that lines $A_1D_1$, $B_1E_1$, and $C_1F_1$ are concurrent.

My proof proceeds in eight steps and uses one well-known lemma about cyclic hexagons. Let $I$ and $\omega$ be the incenter and incircle of $ABCDEF$. Let $I_A$ denote the center of $\omega_A$; define $I_B$, $I_C$, $I_D$, $I_E$, $I_F$ similarly. Step 1. By homothety $A$, $I_A$, $I$ are collinear, as are $B$, $I_B$, $I$; etc. Step 2. Since $FABC$ cyclic $\angle AI_AB = \angle AI_BB$, so $ABI_BI_A$ cyclic and $II_A \cdot IA = II_B \cdot IB$. Then there exists $p$ with $p = II_A \cdot IA$, etc. Step 3. Inverting at $I$ with power $p$, $I_AI_BI_CI_DI_EI_F$ is cyclic. Step 4 (key step I). Consider the transformation $t$ consisting of reflections in lines $I_AI_B$, $I_AA$, $I_AI_F$, $I_AI_D$ in that order. It consists of four reflections in lines through $I_A$, so it is a rotation about $I_A$.

. Step 6 (key step III). Consider the effect of $t$ on $\ell_{AB}$: it is \[\ell_{AB} \stackrel{\overline{I_AI_B}}{\longrightarrow} \overline{AB} \stackrel{\overline{I_AA}}{\longrightarrow} \overline{AF} \stackrel{\overline{I_AI_F}}{\longrightarrow} \ell_{FA} \stackrel{\overline{I_AI_D}}{\longrightarrow} \ell_{AB}\]where $\stackrel{\overline{I_AI_F}}{\longrightarrow} \ell_{FA} \stackrel{\overline{I_AI_D}}{\longrightarrow} \ell_{AB}$ because $t$ is identity. So $\ell_{FA}$ and $\ell_{AB}$ are reflections over $\overline{I_AI_D}$: $A_1$ is on $I_AI_D$ and $D_1$ is too, so lines $I_AI_D$ and $A_1D_1$ coincide.

Step 7. By Brianchon theorem on $ABCDEF$, $\overline{AD}$, $\overline{BE}$, $\overline{CF}$ concur so $\tfrac{AB}{BC} \tfrac{CD}{DE} \tfrac{EF}{FA} = 1$. Step 8. Inverting at $I$ with power $p$, $\tfrac{I_AI_B}{I_BI_C} \tfrac{I_CI_D}{I_DI_E} \tfrac{I_EI_F}{I_FI_A} = 1$ by inversion distance formula. Thus $\overline{I_AI_D}$, $\overline{I_BI_E}$, $\overline{I_CI_F}$ concur; i.e. $\overline{A_1D_1}$, $\overline{B_1E_1}$, $\overline{C_1F_1}$ concur.

$\textbf{Proof :}$ $\textbf{Lemma 1:}$ Let given hexagon $ABCDEF$ where $AB\parallel DE$, $BC\parallel EF$, $CD\parallel FA$ and given that $|AB| + |CD| + |EF| = |BC| + |DE| + |FA|$. Then $AD, BE, CF$ are concurrent. $\textbf{Proof of lemma :}$ Consider midpoints $M_A, M_B, M_C$ of segments $AD, BE, CF$ respectively. Then easy to see that $M_AM_B\parallel AB$, $M_BM_C\parallel BC$ and $M_AM_C\parallel AF$. Also one get that $|M_AM_B| = |AB| - |DE|$, $|M_BM_C| = |BC| - |EF|$ and $|M_AM_C| = |AF| - |CD|$. So $M_A, M_B, M_C$ have to be collinear. And so $M_A = M_B = M_C$ and $AD, BE, CF$ are concurrent. $\Box$ Return to main problem. It's well known that $C_1B_1\parallel AD$. So $C_1B_1\parallel AD\parallel E_1F_1$ and similarly $A_1B_1\parallel D_1E_1$ and $A_1F_1\parallel C_1D_1$. From simple segment computation one get that $|AB| - |BC| + |CD| - |DE| + |EF| - |FA| = |A_1B_1| - |B_1C_1| + |C_1D_1| - |D_1E_1| + |E_1F_1| - |F_1A_1| = 0$. So by lemma 1 we conclude that $A_1D_1, B_1E_1, C_1F_1$ are concurrent. Done

@CantonMathGuy There is a simplification of your proof : 1 step. There exists inversion $i$ which sends $ABCDEF$ to $A_1B_1C_1D_1E_1F_1$. And $AD, BE, CF$ are concurrent, so $(IAD), (IBE), (ICF)$ are coaxial and so $A_1D_1 = i(IAD), B_1E_1, C_1F_1$ are concurrent. 2 step. Let $M_A$ be the midpoint of the biggest arc $BF$ of $(ABC)$ and $M_D$ be the midpoint of the biggest arc $CE$ of $(ABC)$. So $I_DI_A\parallel M_DM_A\parallel $ to external angle bisector between lines $CF, BE$. 3 step. $D_1E_1\parallel CF$ and $C_1D_1\parallel BE$. So $I_DD_1\parallel I_DI_A$ and $D_1\in I_DI_A$. Like the same $A_1\in I_DI_A$. So $A_1D_1, B_1E_1, C_1F_1$ are concurrent.

aleksam wrote: Let $ABCDEF$ be a bi-centric convex hexagon. Let $\omega_{A}, \omega_{B},\omega_{C},\omega_{D},\omega_{E}$ and $\omega_{F}$ be the inscribed circles of the triangles $FAB, ABC, BCD, CDE, DEF$ and $EFA$, respectively. Let $l_{AB}$, be the external tangent of $\omega_{A}$ and $\omega_{B}$; lines $l_{BC}$, $l_{CD}$, $l_{DE}$, $l_{EF}$, $l_{FA}$ are defined analogously. Let $A_1$ be the intersection point of the lines $l_{FA}$ and $l_{AB}$, $B_1, C_1, D_1, E_1, F_1$ are defined analogously. Prove that $A_1D_1, B_1E_1, C_1F_1$ are concurrent. Note that for a cyclic hexagon $ABCDEF$, lines $AD, BE, CF$ are concurrent if and only if $\tfrac{AB}{BC}\cdot \tfrac{CD}{DE}\cdot \tfrac{EF}{FA}=1$. Proving it is easy, just apply trig ceva on $\triangle ACE$ with $AD, BE, CF$ as cevians. Meanwhile, $ABCDEF$ is circumscribed, so we actually have the product relation because of Brianchon's Theorem. Let $I_A, I_B, I_C, I_D, I_E, I_F$ be the respective centres of these six incircles. Note that $\measuredangle AI_AB=\measuredangle AI_BB$ since $F,A,B,C$ are concyclic, so $ABI_AI_B$ is cyclic. Likewise, we obtain $$II_A\cdot IA=II_B\cdot IB=\dots=II_F\cdot IF,$$which together with $ABCDEF$ cyclic yields that $I_AI_BI_CI_DI_EI_F$ is cyclic. Lemma. Lines $A_1D_1$ and $I_AI_D$ coincide. (Proof) Reflect $I_AI_D$ along $I_AI_F$ to meet $FA$ at $F'$ and along $I_AI_B$ to meet $AB$ at $B'$. It suffices to show $I_AF'=I_AB'$ to see $A_1$ lies on $I_AI_D$. Notice that \begin{align*} \angle AI_AF' &= \left(180^{\circ}-\angle IFA \right)-\angle I_FI_AI_D \\ &= \angle IFE+\angle IDE-\angle IFA \\ &= \tfrac{1}{2}\angle D \end{align*}so $\angle AI_AF'=\angle AI_AB'$. Together with $\angle IAF=\angle IAB$ we have $\triangle AI_AF' \cong \triangle AI_AB' \implies I_AF'=I_AB'$ as desired. $\blacksquare$ Finally, note that $A_1D_1, B_1E_1, C_1F_1$ concur if and only if $\tfrac{I_AI_B}{I_BI_C}\cdot \tfrac{I_CI_D}{I_DI_E}\cdot \tfrac{I_EI_F}{I_FI_A}=1$. Now by inversion, $\tfrac{I_AI_B}{I_BI_C}=\tfrac{AB}{BC}\cdot \tfrac{IC}{IA}$. So we have$$\frac{I_AI_B}{I_BI_C}\cdot \frac{I_CI_D}{I_DI_E}\cdot \frac{I_EI_F}{I_FI_A}=\frac{AB}{BC}\cdot \frac{CD}{DE}\cdot \frac{EF}{FA}=1$$as desired. $\blacksquare$ Remark. The main idea was to "guess" the lemma. However, without a diagram as good as that provided by CantonMathGuy, I don't think it's reasonable to to do in time. Pretty hard problem. P.S. How to construct a bi-centric hexagon on geogebra?

anantmudgal09 wrote: P.S. How to construct a bi-centric hexagon on geogebra? First, construct the circumcircle $ \odot (O) $ and the incircle $ \odot (I) $ of $ \triangle ACE, $ then draw the circumcevian triangle $ \triangle DFB $ of $ L $ WRT $ \triangle ACE $ where $ L $ is one of the limiting points of $ \odot (I), \odot (O). $ ______________________________ In my opinion, this problem is not hard (I solved it without drawing a picture). When I saw this problem, the first thought came to my mind is that this problem might not use some deep properties of bicentric hexagon as it's just an olympiad problem. First, I focused on cyclic quadrilateral (it's hard to deal with hexagon...) and then noticed $ AD \parallel B_1C_1 \parallel E_1F_1 $ by a well-known fact --- For a cyclic quadrilateral ABCD, the line connecting the incenter of $ \triangle ABC $ and $ \triangle DBC $ is parallel to the bisector of (AD,BC) , so I guessed $ A_1D_1, _{\dots} $ have same midpoint which led me to use the condition of the existence of incircle.

aleksam wrote: Let $ABCDEF$ be a convex hexagon which has an inscribed circle and a circumcribed. Denote by $\omega_{A}, \omega_{B},\omega_{C},\omega_{D},\omega_{E}$ and $\omega_{F}$ the inscribed circles of the triangles $FAB, ABC, BCD, CDE, DEF$ and $EFA$, respecitively. Let $l_{AB}$, be the external of $\omega_{A}$ and $\omega_{B}$; lines $l_{BC}$, $l_{CD}$, $l_{DE}$, $l_{EF}$, $l_{FA}$ are analoguosly defined. Let $A_1$ be the intersection point of the lines $l_{FA}$ and $l_{AB}$, $B_1, C_1, D_1, E_1, F_1$ are analogously defined. Prove that $A_1D_1, B_1E_1, C_1F_1$ are concurrent. Ok so I am not posting the diagram, extremely difficult to draw smh my head. Like CantonMathGuy, define $I_A, I_B, I_C, I_D, I_E$ and $I_F$. Let $I$ be the incenter of hexagon $ABCDEF$. Let $\Omega$ and $\omega$ be the circumcircle and incircle of hexagon $ABCDEF$. Lemma $1$ : Points $I_A, I_B, I_C, I_D, I_E$ and $I_F$ are concyclic. Proof : Using Fact $5$, considering $\triangle FAB$ and $\triangle CAB$, we know that the points $A, B, I_A, I_B$ lie on a circle with center $M$. Now, we see that points $A, I_A, I$ are collinear since lines $\overline{AI}$ and $\overline{AI_A}$ both bisect $\angle BAF$. Similar holds true if you replace $A$ by $B, C, D \dots$ Now, we see that $II_A \cdot IA = II_B \cdot IB$ is the power of $I$ with respect to circle with center $M$ and radius $MA$, therefore $$II_A\cdot IA=II_B\cdot IB=II_C\cdot IC = II_D \cdot ID = II_E \cdot IE =II_F\cdot IF$$Now inversion about a circle $\Gamma$ centered at $I$ and having radius $\sqrt{II_A \cdot IA}$ inverts cyclic hexagon $ABCDEF$ to a hexagon $I_AI_BI_CI_DI_EI_F$ which must also be cyclic as desired. By Trigonometric form of Ceva's Theorem in $\triangle ACE$, we obtain that $AB \cdot CD \cdot EF = AF \cdot ED \cdot CB$. Using inversion length formula, we see that $AB = I_AI_B \cdot \dfrac{IA \cdot IB}{\sqrt{II_A \cdot IA}}$ and so on, which ultimately gives you that $I_AI_B \cdot I_CI_D \cdot I_EI_F = I_AI_F \cdot I_EI_D \cdot I_CI_B$ which implies that the diagonals $\overline{I_AI_D}, \overline{I_BI_E}, \overline{I_CI_F}$ are concurrent at a point $I_1$ due to Trigonometric Ceva (and perhaps Phantom Points, depending on your argument) Here is the final lemma to this problem. Lemma $2$ : Points $I_A, I_D$ lie on line $\overline{A_1D_1}$ and similarly points $I_B, I_E$ lie on line $\overline{B_1E_1}$ and points $I_C, I_F$ lie on line $\overline{C_1F_1}$, implying that lines $\overline{A_1D_1}, \overline{B_1E_1}, \overline{C_1F_1}$ concur at the point $I_1$. Proof : We only prove that the points $I_A, I_D$ lie on line $\overline{A_1D_1}$, the other claims follow analogously and then we can conclude the problem. Now, we prove that line $\overline{AD}$ is the angle bisector of $\angle F_1A_1B_1$ and of $\angle C_1D_1E_1$ which will prove our claim. We first prove a mini lemma. Lemma $2.1$ : $\ell_{AB} || \overline{CF} || \ell_{DE}, \ell_{BC} || \overline{AD} || \ell_{EF}, \ell_{CD} || \overline{BE} || \overline {AF}$. Proof : If $M$ is the midpoint of arc $AB$ not containing $C$ or $F$ and $\ell_M$ is the tangent from $M$ to $\Omega$, we can see that $\angle FCM = \angle(\ell_M, \overline{FM}) = \angle (\overline{AB}, \overline{FM}) = \angle (\ell_{AB}, \overline{CM})$ which means that $\overline{CF} || \ell_{AB}$ and the others follow analogously. Now, using Lemma $2.1$, we see that if $M_{AB}$ and $M_{DE}$ are the midpoints of arc $AB$ and arc $DE$ not containing $C$ or $F$, then it can be seen that $\angle(\overline{M_{AB}M_{DE}}, \overline{BE}) = \angle(\overline{M_{AB}M_{DE}}, \overline{CF})$ and so it implies that $\overline{M_{AB}M_{DE}}$ is parallel to angle bisectors of $\angle F_1A_1B_1$ and $\angle C_1D_1E_1$, which implies that $\overline{I_AI_D}$ is parallel to angle bisectors of $\angle F_1A_1B_1$ and $\angle C_1D_1E_1$ (this is because with respect to $\angle AID$, lines $\overline{AD}$ and $\overline{M_{AB}M_{DE}}$ are anti-parallel and lines $\overline{AD}$ and $\overline{I_AI_D}$ are anti-parallel) Ultimately, we see the angle bisector of $\angle F_1A_1B_1$ passes through point $I_A$ (property that the angle formed by tangents $\ell_1$ and $\ell_2$ from an external point $D$ to a circle $\gamma$ with center $O$ is bisected by line $\overline{DO}$) and similarly angle bisector of $\angle C_1D_1E_1$ passes through point $I_D$, and since angle bisectors of $\angle F_1A_1B_1$ and $\angle C_1D_1E_1$ both are parallel to line $\overline{I_AI_D}$, it must follow that line $\overline{I_AI_D}$ bisects both these angles, which implies that points $I_A, A_1, D_1, I_D$ are collinear, similarly this means that points $I_B, B_1, E_1, I_E$ and points $C_1, I_C, I_F, F_1$ are collinear, implying that lines $\overline{A_1D_1}, \overline{B_1E_1}$ and $\overline{C_1F_1}$ concur at the point $I_1$, as desired.