Let $C'$ be a symmetrical point to $C$ with respect to $I$, $A'$ lie on line $CA$ such that $CA' = 2AB$ and $B'$ lie on line $CB$ such that $CB' = 2AB$. $CA + CB = 3AB$ imply $CD = AB$, $BB' = DA$ and $AA' = EB$. We see, that $C'B' = C'A' = 2IE$ and $C'K = C'L = CE = AB$. Triangle $C'B'B$ is congruent to $ADK$ and $C'A'A$ to $BLE$. Thus $C'B = AK$ and $C'A = LB$. So triangle $C'BK$ is congurent to $ABK$ and $C'AL$ to $ABL$. Thus $\angle KC'B = \angle KAB$ and $\angle LC'A = \angle LBA$. It means that $AKBC'$ and $ALBC'$ are cyclic quadrilaterals thus $ALKB$ is cyclic quadrilateral.

Do you infer that $C'$ is the excircle of $ABC$?? Edit: Yes. I see why $C'$ is the excircle of $ABC$. From the similar triangles $AIE$ and $AC'A'$ we find that $C'A'\perp AC$. Similarly for $C'B'\perp BC$. But, $A'$ and $B'$ are the poits of tangency of the excircle. ($CA'=CB'=\frac p 2$) Sorry for my useless post.

xirti wrote: Do you infer that $C'$ is the excircle of $ABC$?? Edit: Yes. I see why $C'$ is the excircle of $ABC$. From the similar triangles $AIE$ and $AC'A'$ we find that $C'A'\perp AC$. Similarly for $C'B'\perp BC$. But, $A'$ and $B'$ are the poits of tangency of the excircle. ($CA'=CB'=\frac p 2$) Sorry for my useless post. i think here we can make it better. from CA' = 2AB = $\frac{AB + BC + CA}{2}$, then C' is the external center of triangle ABC. thus I, A, B, C' are lie on circle with diameter IC'. another side, we have K, L, C' are the symmetrycal points of D, E, C then C'K, C'L are the tangents of (I), then K, L also lie on circle with diameter IC'.

I think Michal Marcinkowski has some typos on his notatations or my figure is wrong.......

This problem is certainly on the ISL from 2005. Question G1 I believe.

PP. Let $ABC$ be an acute triangle. Its incircle $w=C(I,r)$ touches it at the points $D\in BC$ , $E\in CA$ , $F\in AB$ . Denote: $\blacktriangleright$ the centroid $G$ of the triangle $ABC$ ; $\blacktriangleright$ the A-exincircle $w_a=C(I_a,r_a)$ , the B-exincircle $w_b=C(I_b,r_b)$ , the C-exincircle $w_c=C(I_c,r_c)$ ; $\blacktriangleright$ the Nagel's point $N\in AD_1\cap BE_1\cap CF_1$ ; $n_a=AD_1$ , $n_b=BE_1$ , $n_c=CF_1$ ; $\blacktriangleright$ the points $D_1\in BC\cap w_a$, $E_1\in CA\cap w_b$ , $F_1\in AB\cap w_c$ ; $\blacktriangleright$ the reflections $D'$ , $E'$ , $F'$ of the points $D$ , $E$ , $F$ with respect to the point $I$ . Lemma 1. $IG\perp BC\Longleftrightarrow b=c\ \ \vee\ \ b+c=3a$ . Proof. $IG\perp BC$ $\Longleftrightarrow$ $IB^2-IC^2=GB^2-GC^2$ $\Longleftrightarrow$ $\frac{ac(p-b)}{p}-\frac{ab(p-c)}{p}=$ $\frac 19\left\{\left[2\left(a^2+c^2\right)-b^2\right]-\left[2\left(a^2+b^2\right)-c^2\right]\right\}$ $\Longleftrightarrow$ $3a(c-b)=c^2-b^2$ $\Longleftrightarrow$ $b=c\ \ \vee\ \ b+c=3a$ . Remark. $N\in (IG)$ and $GN=2\cdot GI$ . Therefore, $IN\perp BC$ $\Longleftrightarrow$ $b+c=3a$ . Lemma 2. $\boxed {a n^2_a+a(p-b)(p-c)=p(b^2+c^2)-bc(b+c)}\ \ (1)$ ; $\frac{NA}{a}=\frac{ND_1}{p-a}=$ $\frac{n_a}{p}\ \ (2)$ ; $D'\in AD_1$ and $\frac{D'A}{p-a}=\frac{D'D_1}{a}=\frac{n_a}{p}\ \ (3)$ a.s.o. Proof. Apply the Stewart's relation for the Nagel-ray $[AD_1$ in the triangle $ABC$ and obtain the relation $(1)$ . Apply the van Aubel's relation $\frac{NA}{ND_1}=\frac{F_1A}{F_1B}+\frac{E_1A}{E_1C}$ and obtain the relation $(2)$ . $D'\in AD_1\cap BC$ $\Longleftrightarrow$ $\frac{D_1D'}{D_1A}=\frac{2r}{h_a}$ $\Longleftrightarrow$ $\frac{D_1D'}{D_1A}=\frac ap$ $\Longleftrightarrow$ $\frac{D_1D'}{a}=\frac{D_1A}{p}=$ $\frac{D'A}{p-a}=\frac{n_a}{p}$, i.e. the relations $(3)$ . Remark. $\frac{AD'}{p-a}=\frac{ND_1}{p-a}=\boxed {\frac{ND'}{2a-p}=\frac{NA}{a}=\frac{n_a}{p}}\ \ (4)$. An equivalent and complete enunciation of the proposed problem. Let $ABC$ be an acute triangle for which $b\ne c$ . Then $B,C,E',F'$ are cyclically $\Longleftrightarrow$ $b+c=3a$ $\Longleftrightarrow$ $IN\perp BC$ $\Longleftrightarrow$ $N\equiv D'$ and in this case $I,B,C,E',F'$ belong to the circle $C(M,MI)$ , where the point $M$ is the second intersection between the line $AI$ and the circumcircle of the triangle $ABC$ . Proof (indication). The points $B,C,E',F'$ are cyclically $\Longleftrightarrow$ $NE'\cdot NB=NF'\cdot NC$ a.s.o. Example. $a=3$, $b=4$, $c=5$. Remark. $N\in w$ $\Longleftrightarrow$ $(b+c-3a)(c+a-3b)(a+b-3c)=0$ $\Longleftrightarrow$$N\in \{D',E',F'\}$ $\Longleftrightarrow$ $p^2+4r^2=16Rr$ .

We draw a circumcircle of the triangle $ABC$. Now let $BI$ meet the circumcircle at the point $M$ (for $M$ is different from $B$), so we have $MC=MA$ since $\angle MAC = \angle MCA= \angle MBC = \angle MBA$. Thus we have $MA=MI=MC$, since $\angle MIA= \angle IAB + \angle IBA=\angle IAM= \angle IAC \angle CAM$. Therefore we have $\angle MIA = \angle MAI$ similarly we have $\angle MIC = \angle MCI$. So resulting we have $MI=MC=MA$. Now lets draw a parallel line to the line $DB$ from point $M$ and let the line meet $IL$ at $N$ so $DB||MN$. Then we obtain a congrunecy between the triangles $MNL$ and $CPM$ ( $P$ is a midpoint of the side $AC$) these facts imply to $ML=MC$. Similarly we get $MK=MA$. Then we can easily say that the points $A,K,I,L,C$ all lie in a circle with centre $M$. davron

See the my first reply where I offered a equivalent and complete enunciation of the proposed problem !

my solution is a "little" ugly because i'm not too good at geometry as to define "miraculous" points, however i consider it a little more natural than any of the above i think it is normal to consider most of the times one's solution as the most natural. it is as follows... let F be the tangency point of the incircle with AB. a simple angle chasing shows that $\angle EFK=\angle AEK=\frac{\theta}{2}$ and $\angle FEK=\angle AFK=\frac{\beta}{2}$. since KL=DE and they are parallel it follows that CI is the perpendicular bisector of KL, which meets the perpendicular bisector of AB at the midpoint of the minor arc AB. but it is a well-known fact that this point is the circumcenter of AIB, thus by symmetry it is enough to prove that AKIB is cyclic. we'll have to prove that $\angle IKM=\frac{\beta}{2}$, or it is equivalent to prove that $\angle AMB=90-\frac{\beta}{2}=\angle FDB$, so we have to prove that $AK$ is parallel to $DF$, which is equivalent to prove that $\angle KAF=90-\frac{\beta}{2}$. let $\angle KAC=\frac{\alpha-\rho}{2}$ and $\angle KAF=\frac{\alpha+\rho}{2}$. we want to prove that $\theta=\rho$. by trig form of ceva's theorem in triangle AEF we have that $\frac{\sin^2(\frac{\beta}{2})\cdot \sin (\frac{\alpha-\rho}{2})}{\sin^2(\frac{\theta}{2})\cdot \sin (\frac{\alpha+\rho}{2})}=1$. a little algebra, the help of briggs formulas, and the condition help us proving that $\frac{\sin^2(\frac{\beta}{2})\cdot \sin (\frac{\alpha-\theta}{2})}{\sin^2(\frac{\theta}{2})\cdot \sin (\frac{\alpha+\theta}{2})}=1$. so we have that $\frac{\sin (\frac{\alpha-\rho}{2})}{\sin (\frac{\alpha+\rho}{2})}=\frac{\sin (\frac{\alpha-\theta}{2})}{\sin (\frac{\alpha+\theta}{2})}$ and it is easy to conclude that $\theta=\rho$ as desired.

The problem proposed by Greece

Check this out campos!!! I bet you'll like this Drawing a good figure I conjectured that $K$ and $L$ lie on the circumcircle of triangle $ABI$. I'll just prove that $L$ lies on this circle, and the result will follow analogously for $K$. The first proof uses some trigonometry. We want to prove that quadrilateral $ABLI$ is cyclic, and since $\angle ABI=B/2$, it suffices to show that $\angle ALI=B/2$. Observe that in triangle $AEL$ we have \[\tan ALE=\frac{AE}{EL}=\frac{s-a}{2r}\] where $s$ is the semiperimeter of the triangle and $r$ is the inradius. Now we use the condition $a+b=3c$ for finding r: \[r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}=\sqrt{\frac{(s-a)(s-b)c}{2c}}=\sqrt{\frac{(s-a)(s-b)}{2}}\] whence \[\tan ALE=\frac{s-a}{2}\sqrt{\frac{2}{(s-a)(s-b)}}=\sqrt{\frac{2(s-a)^{2}}{4(s-a)(s-b)}}=\sqrt{\frac{s-a}{2(s-b)}}\] But we know that \[\tan \frac{B}{2}=\sqrt{\frac{(s-c)(s-a)}{s(s-b)}}=\sqrt{\frac{c(s-a)}{2c(s-b)}}=\sqrt{\frac{s-a}{2(s-b)}}\] Thus $\tan ALI=\tan B/2$, that is, $\angle ALI=B/2$, and we are done. Now I continue with a second synthetic proof I found. Let $P$ be a point on the segment $AC$ such that $AB=AP$. Note that $EP=c-(s-a)=c+a-s=s-b$. We claim that triangles $AEI$ and $LEP$ are similar. Indeed, we have $\angle AEI = \angle LEP= \pi/2$ and \[\frac{AE}{LE}=\frac{EI}{EX}\Leftrightarrow \frac{s-a}{2r}=\frac{r}{s-b}\Leftrightarrow r=\sqrt{\frac{(s-a)(s-b)}{2}}\] Hence $\angle APL=\pi/2-A/2$. But since triangle $ABP$ is isosceles, $\angle APB= \pi/2-A/2$; so $\angle APB=\angle APL$ and therefore $B$, $L$ and $P$ are collinear. Now we have $\angle BAI=\angle ILP=A/2$, that is, quadrilateral $ABLI$ is cyclic, and we are done.

nice, gabriel! congrats, and i must admit it is much more natural and nice than mine!

Sketch-proof of method using polars. Let $F$ be the point of tangency with the incircle and $AB$. It can be shown that $EF$ is bisected by $KL$ (a short sine-rule bash). But the midpoint of $EF$ is the inverse of $A$ wrt the incircle, and similarly with $DF$. Since lines invert to circles, with $K$,$L$ self inverse, we have $A$,$B$,$K$,$L$ concyclic as required.

Here is a sketch of a solution: Let $BL\cap AC=E''$, $AK\cap BC=D''$, line $l$ touches the incircle at $L$, $l\cap AB=F'$ and $l\cap BC=D'$. $AK\cap BL=X$. $l//AC$ since $\angle AEI=\angle F'LI=90$. The $B$-excircle of $ABC$ touches $AC$ at $E''$. Consider a homothety center $B$ that maps $l$ to $AC$. $\frac{BL}{BE''}=1-\frac{br}{A}=\frac{d(B,AC)-2r}{d(B,AC)}$. Use mass points to determine $\frac{E''X}{XB}$ in terms of the sides. Use: $E''C=s_{ABC}-a$, $D''C=s_{ABC}-b$ Use stewart's theorem to determine the length of $BE''$ and $D''A$. Compute the product, $XK*XA-XL*XB$. [use $b+a=3c$] If this turns out to be $0$, then $ADLB$ is cyclic by the converse of the power of a point theorem. Note: Perhaps the excircle can be used with angle information. I will look into that.

Okay I finally figured out how to use that pesky $3c=a+b$... -- Denote $w_{B}$ as the $B$-excircle of $ABC$, $AC\cap BL=E''$, $w_{B}\cap AB=T$, $l$ is the line tangent to $C(I,r)$ at $L$. Since $EL\perp AC$ and $EL\perp l$ and $l//AC$. Let $H$ be a homothety center $B$ that maps $l$ to $AC$. Since $C(I,r)$ is tangent to $l$, $AB$ and $AC$, We know that $H(C(I,r))$ is tangent $H(l)=AC$, $H(AB)=AB$, and $H(BC)=BC$, considering betweeness, $H(C(I,r))=w_{B}$. It follows that $w_{B}$ touches $AC$ at $E''$, so $TA=AE''=s-c=\frac{1}{2}*(a+b-c)=c$. Hence, $AE''=AB$, so $\angle LBA=\frac{1}{2}*(\pi-\angle A)$. By simple angle chasing, $\angle AIB=\frac{1}{2}*(\pi+\angle A)$. Then $\angle AIB+\angle LBA=\pi$. Hence, $L$ lies in the circumcircle of $AIB$. By symmetry, $K$ lies on that circle as well. It follows that $AKLB$ is cyclic.

I found a new solution... let $ M$ be the midpoint of arc $ AB$. we'll prove that $ M$ is the circumcenter of $ AKILB$. we know that $ MA=MI=MB=2R\sin \frac{A}{2}$, so it's enough to prove that $ MK=MI$. we'll prove first that $ r=4R\sin^2\frac{C}{2}$. this is equivalent to $ \sin \frac{A}{2}\sin\frac{B}{2}=\sin\frac{C}{2}$, or equivalently $ s-c=c$, which is true from the hyphotesis. then, note that $ \angle MIK=\angle CID= 90-\frac{C}{2}$. let $ N$ be the projection of $ M$ on $ IK$. then, $ IN=IM\cdot \sin\frac{A}{2}=2R\sin^2\frac{C}{2}=\dfrac{IK}{2}$. then, $ MN$ is a median too, thus, $ MI=MK$, as we wanted to prove.

Let $ P = AK\cap BC$, $ Q = BL\cap CA$. It's well-known that $ P$ is the tangency point of $ A$-excircle with $ BC$ and $ Q$ the tangency point of $ B$-excircle with $ CA$. It implies $ AQ = CE = AB$, $ BP = AB$. We have $ \angle QLK = \angle QLE + \angle ELK = \angle CAB/2 + \angle BCA/2 = \angle KAB$. Hence, quadrilateral $ AKLB$ is cyclic.

We know that $ AK$ cuts $ BL$ at $ N$, with $ N$ is Nagen point of $ \delta ABC$. We’ll prove that $ N$ lie on $ (I)$ and $ N, I, T$ is collinear Proof by using vector, easy!: proof that $ \vec{IN}+\vec{IT} = \vec{0}$ Hence, $ NK.NA = NT^{2} = NL. NB$ So $ A, B, K, L$ lie on a circle PS: If using vector to solve, we can find all value of $ k$ such that: if $ BC+CA = k.AB$ then $ A, B, K, L$ lie on a circle

Denote by $ M$ the midpoint of $ AB$, let $ T$ be the circumcenter of quadrilateral $ CEID$ and let $ S$ be the circumcenter of $ \triangle ABI$. Clearly $ S$ lies on the circumcircle of $ \triangle ABC$ and $ C$, $ I$, $ S$, $ T$ are collinear. We'll now prove that $ S$ is the circumcenter of $ \triangle IKL$. The triangles $ \triangle IKL$ and $ \triangle IDE$ are congruent, moreover they are symmetric wrt the perpendicular of $ SC$ through $ I$. Thus, it's enough to prove that $ TI = SI$, i.e. $ CI/IS = 2$. But $ \angle CEI = \angle AMS = 90^ o$ and $ \angle ICE = \angle MAS = \gamma /2$. Thus, triangles $ \triangle ASM$ and $ \triangle CIE$ are similar. But $ CE = (AC + BC - AB)/2 = AB$. Thus $ CE = 2AM$ and $ CI/IS = 2$ follows.

Notice that $CD=CE=AB$. Let $AK$ hit $BC$ at $X$. Let $BL$ hit $AC$ at $Y$. Then $X$ and $Y$ are the contact points of the $A$ and $B$-excircles with $BC$ and $AC$. Then, notice that $\triangle ABX$ is isosceles and $\triangle ABY$ is isosceles. It suffices to show that $\angle YLK=\angle XAB=\frac{180^{\circ}-\angle B}{2}$. WLOG, suppose that $AC\le BC$. Then if $DE$ intersects $AB$ at some point $M$ to the left of $AB$ and $BY$ intersects $DE$ at $P$ then we have that \begin{align*} \angle YLK&=\angle YPE \\ &= \angle EMA+\angle YBA\\ &=\angle CDE-\angle B+\frac{180^{\circ}-\angle A}{2} \\ &=\frac{180^{\circ}-\angle C}{2}-\angle B+\frac{180^{\circ}-\angle A}{2} \\ &=\frac{180^{\circ}-\angle B}{2} \end{align*}so we are done. $\blacksquare$

We use barycentric coordinates, with $ABC$ as the reference triangle. We have the following points: \begin{align*} I &= (3a:3b:a+b) \\ D &= (0:a+b:2a-b) \\ E &= (a+b:0:2b-a) \\ \end{align*}Now, we need to find $K$. Since $I$ is the midpoint of $DK$, we know that $K=2I-D$. Some computation gives $$K =\left(9a^2 : 5ab - 2a^2 - 2b^2 : ab - a^2+2b^2\right).$$Similarly, we have $$L = \left(5ab - 2a^2-2b^2 : 9b^2 : ab + 2a^2 - b^2\right).$$Next, consider $(ABK)$. If we can show that $L$ also lies on this circle, we are done. First, we evidently have $u=v=0$. We need $w$, and this is just more computation: $$(b-2a)\left(-a^2+2ab+a^2+ab\right)-9ab^2=-w(6a+6b) \rightarrow 6ab(a+b)=w(6a+6b) \rightarrow w=ab.$$Clearly, $L$ also lies on this circle, which finishes the problem. $\blacksquare$

Let $M$ the midpoint of the smaller arc $\widehat{AB}$. It is well-known that $MA=MB=MI$. From Ptomely's theorem on quadrilateral $CAMB$ we have: \begin{align*} MA \cdot BC + MB \cdot CA = AB \cdot CM \\ MI (BC+CA) = AB \cdot CM \\ 3MI = CM \end{align*}Define $T$ to be a reflection of $M$ over $I$, then from before the established result, we have $CT=TI=IM$. Note that $T$ is the centre of $\odot(IDCE)$, which is cyclic deltoid. Therefore by symmetry, we have that $\triangle DTI$ and $\triangle ITE$ are congruent isosceles triangles. The key observation is that $\angle DIT = \angle TIE = \angle LIM = \angle KIM$ and $TI=IM$ combined with $ID =IE =IK =IL$. This implies that $\triangle DTI =\triangle ETI =\triangle LMI = \triangle KMI$, which gives us that $LM=MI=MK=MA=MB$. This means that points $A,L,I,K,M$ all lie on the same circle centred at the point $M$ with radius $MI$. This solves the problem.

Attachments:

Solved with support from Taco12. We use barycentric coordinates with reference triangle $ABC$. It is well known that $$I=\left(\frac{a}{a+b+c},\frac{b}{a+b+c},\frac{c}{a+b+c}\right)=\left(\frac{3a}{4(a+b)},\frac{3b}{4(a+b)},\frac{1}{4}\right),$$$$D=\left(0,\frac{a+b-c}{2a},\frac{a-b+c}{2a}\right)=\left(0,\frac{a+b}{3a},\frac{2a-b}{3a}\right),$$$$E=\left(\frac{a+b-c}{2b},0,\frac{-a+b+c}{2b}\right)=\left(\frac{a+b}{3b},0,\frac{2b-a}{3b}\right).$$$K$, being the reflection of $D$ over $I$, is equivalent to $$2I-D=\left(\frac{3a}{2(a+b)},\frac{3b}{2(a+b)}-\frac{a+b}{3a},\frac{1}{2}-\frac{2a-b}{3a}\right).$$$L$, being the reflection of $E$ over $I$, is equivalent to $$2I-E=\left(\frac{3a}{2(a+b)}-\frac{a+b}{3b},\frac{3b}{2(a+b)},\frac{1}{2}-\frac{2b-a}{3b}\right).$$We can unhomogenize $K$ to be $$\left(9a^2 : 9ab-2(a+b)^2 : (a+b)(2b-a)\right).$$We can unhomogenize $L$ to be $$\left(9ab-2(a+b)^2 : 9b^2 : (a+b)(2a-b)\right).$$The equation of a circle is $$-a^2yz-b^2zx-c^2xy+(ux+vy+wz)(x+y+z)=0,$$and since $A$ and $B$ are on it, we have that $$0+(u(1)+0+0)(1)=0\Rightarrow u=0,$$$$0+(0+v(1)+0)(1)=0\Rightarrow v=0.$$Therefore, if we can show that both $K$ and $L$ satisfy the equation with the same $w$, we are done. We can see that $$w=\frac{a^2yz+b^2zx+c^2xy}{z(x+y+z)},$$so all that's left is verifying that $w$ is indeed equal for both points. This turns out to be true, and I will include the computation below. QED. $$\frac{a^2(9ab-2(a+b)^2)(a+b)(2b-a)+b^2(9a^2)(a+b)(2b-a)+c^2(9a^2)(9ab-2(a+b)^2)}{(a+b)(2b-a)\left(9a^2+9ab-2(a+b)^2+(a+b)(2b-a)\right)}=\frac{a^2(9b^2)(a+b)(2a-b)+b^2(9ab-2(a+b)^2)(a+b)(2a-b)+c^2(9ab-2(a+b)^2)(9b^2)}{(a+b)(2a-b)\left(9ab-2(a+b)^2+9b^2+(a+b)(2a-b)\right)}$$$$\frac{a^2(7b-2a)(a+b)^2(2b-a)+(a+b)^2(a^2)(9ab-2(a+b)^2)}{(a+b)(2b-a)(6a)(a+b)}=\frac{b^2(7a-2b)(a+b)^2(2a-b)+(a+b)^2(9ab-2(a+b)^2)(b^2)}{(a+b)(2a-b)(6b)(a+b)}$$$$\frac{a^2(a+b)^2(6b)(2b-a)}{a(2b-a)}=\frac{b^2(a+b)^2(6a)(2a-b)}{b(2a-b)}$$$$\frac{a^2b(2b-a)}{a(2b-a)}=\frac{b^2a(2a-b)}{b(2a-b)}$$$$ab=ab$$

Replace $K$ and $L$ with $D'$ and $E'$. Add $F$, the $C$-intouch point, as well as its antipode $F'$ wrt the incircle. Also add the midpoints $M,N$ of $\overline{BC},\overline{AC}$ resp. and let $P,Q$ be the $A$-extouch and $B$-extouch points. Also let $s,a,b,c$ be defined as usual. It is easy to see that $c=AB=AQ=BP$. Furthermore, $\overline{MN}$ is tangent to the incircle by Pitot, in fact at $F'$. Finally, it is well-known that $D'$ lies on $\overline{AP}$ and $E'$ lies on $\overline{BQ}$. We now have the following key claim. Proof: $F'$ lies on $\overline{AP}$ and $\overline{BQ}$.

Claim: Symmetry: $D$ and $F$ are symmetric with respect to $B$, which implies that $D'$ and $F'$ are, as are $A$ and $P$, so $F'$ lies on $\overline{AP}$ follows from the fact that $D'$ lies on it. Similarly, $F'$ lies on $\overline{BQ}$. Note

To finish we employ an angle chase: $$180^\circ-\angle AD'E'=\angle F'D'E'=\angle FDE=\frac{\angle FIE}{2}=90^\circ-\frac{\angle FAE}{2}=\angle ABE',$$where the last line follows from $AB=AQ$, implying the desired result. $\blacksquare$

The length condition is just $s=2c$. Let $D'$ be the excircle touchpoint to $BC$ (or the reflection of $D$ across the midpoint of $BC$). It is well known that $A,K,D'$ are collinear. However, $$BD'=s-c=c,$$so $\angle BAD'=\angle BD'A$. Let $\angle ABC=2\beta$. Then, $\angle BAD'=\angle BD'A=90-\beta$. However, $\angle BID=90-\beta$ as well, so $AKIB$ is cyclic. Similarly, $AILB$ is cyclic. Hence, $ABLK$ is cyclic

Let $D'$ be the $A$-excentral touchpoint, thus the length condition implies $AB=BD'$; also, $A, K, D'$ are collinear. The rest is just an angle chase. Observe \begin{align*} \angle DKL &= \angle D'KD + \angle LKD \\ &= 90^\circ - \angle KD'D + 90^\circ - \angle ELD \\ &= 90^\circ - (90^\circ - \frac B2) + 90^\circ - 90^\circ - \frac C2 \\ &= \frac B2 + \frac C2 \end{align*}On the other hand, $\angle BAD' = 90^\circ - \frac B2$, implying $\angle ABL = 90^\circ - \frac A2$ symmetrically (the length condition still applies.) This is enough to prove the concyclicity.

Let the incircle touch side $AB$ at point $F,$ and let $M$ be the reflection of $F$ with respect to $I.$ We further claim that $ABKLI$ is cyclic. The first step is to invert with respect to the incircle. The point $A$ gets mapped to the midpoint $X$ of $EF,$ and the point $B$ gets mapped to the midpoint $Y$ of $DF.$ Thus it suffices to prove that $K,L,X,Y$ are collinear. For now, we will only focus on the point $X$; the analysis for the point $Y$ is symmetrical to that of $X.$ A rephrasing is to prove that if $EF$ intersects $KL$ at $X,$ then $X$ is the midpoint of $EF.$ To do so, we will use projective geometry. If $X$ were indeed the midpoint of $EF,$ then we'd have $(E,F;X,P_{\infty}) = -1.$ Projecting through the point $L$ and noting that $LM \parallel EF,$ we get $(E,F;K,M) = -1.$ Thus it suffices to prove that $EFKM$ is harmonic. To do so, since the tangents at $E,F$ intersect at $A,$ we must prove that $A,K,M$ are collinear. It turns out that $M$ is the Nagel point of $\triangle ABC,$ and this can easily be proven using barycentric coordinates and the given condition on the triangle.

With a good diagram, we see that I lies on this as well. Indeed, BID=90-B/2 $\implies$ KIB=90+B/2. On the other hand, the fact that 3AB=AE+BD+2CD $\implies$ 4AB=2AB+2CD=2s leads to CD=s-c=c. Since we can't do much with BAK, it's well known that extending this intersects at D', the reflection of D over midpoint of BC (motivating because in these concise problems, there will probably be some construction). Then BD'=BA and we get that BAD'=BDA', whence BAK=BAD'=90-B/2, so ABIK is cyclic, and analogously so is ABLI, so ABLK is cyclic, as desired. $\blacksquare$

$(CI)$ and $(II_C)$ are reflections about $I$, so $CIDE$ and $II_CAB$ cyclic implies $ABKL$ cyclic.

We invoke barycentric coordinates with $\triangle ABC$ as our reference triangle. We can see that our length condition can be used in that $I = (3a:3b:a+b)$. Notice now that $D = (0:a+b:2a-b)$ and $E = (a+b:0:2b-a)$ it can now be computed that $K = (9a^2:2a^2+5ab+2b^2:-a^2+ab-2b^2)$ and similarly $L = (2a^2 + 5ab+2b^2:9b^2:-2a^2+ab-b^2)$. Bash the rest from here. I believe in you. It's 12:13 so I must go to bed.

Let $M$ be the midpoint of $\overline{CI}$ and let $P$ be the midpoint of arc $AB$. Then, by ASA congruence we have $\triangle CME \cong \triangle APB$. In particular, we have $PI = IM = EM = DM$. Reflecting segments $\overline{EM}$ and $\overline{DM}$ over the perpendicular bisector of $\overline{MK}$ gives us segments $\overline{KP}$ and $\overline{LP}$ respectively, so \[ PI = IM = EM = DM = PK = PL,\]hence the result follows.

Here is a solution using complex numbers. Let the incircle touch $AB$ at $F$. Let's take $(DEF)$ as the unit circle. $$a=\frac{2ef}{e+f} , b=\frac{2df}{d+f}, c=\frac{2de}{d+e}$$ We have $|CE|=|CD|=\frac{|AB|+|AC|+|BC|}2-|AB|=\frac{4|AB|}2-|AB|=|AB|$ So, $$|a-b|=|c-e|$$$$(a-b).(\overline{a}-\overline{b})=(c-e).(\overline{c}-\overline{e})$$$$(\frac{2ef}{e+f}-\frac{2df}{d+f}).(\frac{2}{e+f}-\frac{2}{d+f})=(\frac{2de}{d+e}-e).(\frac{2}{d+e}-\frac{1}{e})$$$$\frac{2f^2.(e-d)}{(d+f).(e+f)}.\frac{2.(d-e)}{(d+f).(e+f)}=\frac{e.(d-e)}{d+e}.\frac{(e-d)}{(d+e)e}$$$$\frac{4f^2}{(d+f)^2.(e+f)^2}=\frac{1}{(d+e)^2}$$$$(\frac{2f}{(d+f).(e+f)})^2=(\frac{1}{d+e})^2$$If $\frac{2f}{(d+f).(e+f)}=\frac{1}{d+e} \Rightarrow f^2-de-df+de=0 \Rightarrow (f-d).(f-e)=0$ which is clearly wrong. So we have $\frac{2f}{(d+f).(e+f)}=-\frac{1}{d+e} \Rightarrow \boxed{f^2+3df+3ef+de=0}(\star)$ Since $K$ and $L$ are reflections of the points $D$ and $E$ with respect to $I$, $$k=-d,l=-e$$ $ABKL$ is cyclic $\Leftrightarrow$ $\frac{(b-a).(k-l)}{(k-a).(b-l)} \in \mathbb{R}$ $$\frac{(b-a).(k-l)}{(k-a).(b-l)}\overset{?}{=}\frac{(\overline{b}-\overline{a}).(\overline{k}-\overline{l})}{(\overline{k}-\overline{a}).(\overline{b}-\overline{l})}$$$$\frac{(\frac{2df}{d+f}-\frac{2ef}{e+f}).(e-d)}{(-d-\frac{2ef}{e+f}).(\frac{2df}{d+f}+e)}\overset{?}{=}\frac{(\frac{2}{d+f}-\frac{2}{e+f}).(\frac{1}{e}-\frac{1}{d})}{(-\frac{1}{d}-\frac{2}{e+f}).(\frac{2}{d+f}+\frac{1}{e})}$$$$\frac{\frac{2f^2.(d-e).(e-d)}{(d+f)(e+f)}}{\frac{(de+df+2ef).(de+ef+2df)}{(d+f).(e+f)}}\overset{?}{=}\frac{\frac{2.(e-d).(d-e)}{(d+f).(e+f).de}}{\frac{(2d+e+f).(d+2e+f)}{(d+f).(e+f).de}}$$$$\frac{f^2}{(de+df+2ef).(de+ef+2df)}\overset{?}{=}\frac{1}{(2d+e+f).(d+2e+f)}$$$$f^2.(f^2+3df+3ef)+f^2.(2d^2+2e^2+5de)\overset{?}{=}f^2.(2d^2+2e^2+5de)+de.(de+3df+3ef)$$$$f^2.(f^2+3df+3ef)\overset{?}{=}de.(de+3df+3ef)$$By $(\star)$ $\Rightarrow$ $-def^2\overset{?}{=}-def^2$ Which is true as desired. $\blacksquare$

Extend $AK$ to meet $BC$ at $M$ and extend $BL$ to meet $AC$ at $N$. By taking homotheties from the incircle to the $A$ and $B$-excircles, we get that $M$ and $N$ are the $A$ and $B$-extouch points, respectively. This gives us that $AN=BM=s-c=2AB-AB=AB$ using the length condition. The isosceles triangles $\triangle ABM$ and $\triangle ABN$ give us enough information to angle chase. We can calculate that \begin{align*} \angle KLI &= \angle IED = \angle ICE = \frac12 \angle C \\ \angle ELN &= 90 - \angle ANB = 90 - (90-\frac12 \angle A) = \frac12 \angle A. \end{align*}Thus we get that $\angle KLB = 180 - \angle KLI - \angle ELN = 180-\frac12 \angle C - \frac12 \angle A = 90+\frac12 \angle B$. $\angle KAB = \angle MAB = 90 - \frac{ 1}{2}\angle B$ so $\angle KLB + \angle KAB = 180$, implying that $A$, $B$, $K$, and $L$ are concyclic, as desired.

(bad formatting because copied from my solution on mathdash) A center the prolbme, so we get AC + AB = 3*BC, incircle is tangent to BC,AC,AB, at D,E,F. K,L are reflections of E,F. it is well knwon and easy to prove that BK meets AC at the reflection of E over the midpoint of AC (consider homothety from incircle to excircle, K goes to tangency point of excircle, which is precisley the reflecfion of E over the midpoint of AC). let BK meet AC at X, then we know that CX = AE = s - a = AB, so CXB is isoceles. as a result, we can conclude CBK = 90 - c/2 and BCL = 90 - b/2. now we desire to prove that BKL + BCL = 180. we compute BKL as BKD + DKL. BKD = 180 - DBK - BDK = 180 - (90 - c/2) - DEK = 90 + c/2 - (90 - DFE) = 90. now DKL = DFL = 90 - DLF = 90 - DEF = b/2. thus BKL = 90 + b/2, so BKL + BCL = 180.