Draw $U$, the midpoint of arc $BC$. Note that $I$ is the orthocenter of $STU$, which implies $AU \perp ST = KL$. However, this implies, since $AU$ is both angle bisector and altitude of $AKL$, that $AK = AL$. At this point, we shall prove that $I$ is the reflection of $A$ with respect to $ST$, which is true since $\angle AST = \angle TSB$, $\angle CTS = \angle STA$, and further by Fact 5 / Incenter-Excenter Lemma, that $AT = TI$, $AS = SI$, which together implies congruence between $\triangle ATS$, $\triangle ITS$, and reflection. Thus $IK = AK = AL = IL$ and we are done.

My sol from exam: 1.$BSKI$ cyclic $\implies{AL // IK}$ 2.$CTLI$ cyclic $\implies {AK // LI}$ 3.$KL$ perpendicular to $AI$ rest is easy.

Any solution is better than no solution The key point is that TKIB is cyclic Notation: (XY) stands for small arc XY Let <BAI=<IAC=x, <ACI=<ICB=y, <CBI=<IBA= z By angles triangle sum 2x+ 2y+2z=180^o => x+y+z =90^o <STC=(SC)/2= <SBC=<ABI=x => TKIB is cyclic So < KIT= (KB)/2= <KBT=(AT)/2= <ACT=z <TIB = <ICB + <IBC = y+ z From angles triangle sum in triangle KIB we get that < BKI= 2x = <KAL = > KI // AL Similarly IL // AK and combining KI // AL we have that ALIK is parallelogram <TBI = <TBK + <KAB = y + z = <TIB => TB = TI And because TB =TA (because arcs AT, TB are equal due to C- angle bisector) We get that AT = TI so T lies on perpendicular bisector of AI Similarly we get that S lies on the same perpendicular bisector So combining those , we conclude that TS is perpendicular bisector of AI So TS _I_ AI , which means KL _I_AI So parallelogram ALIK becomes a rhombus

Let $D$ be the intersection point of $AI$ and $KL$ $B;K;A$ colinear $C;L;A$ colinear $B;I;S$ colinear $C;I;T$ colinear $T;K;L;S;D$ colinear $\measuredangle CBS=\measuredangle SBA=\alpha$ $\measuredangle ACT=\measuredangle TCB=\beta$ $\measuredangle KAD=\measuredangle DAL=90-(\alpha+\beta)$ Claim:$BTKI$ is cyclic Proof: $BTSC$ cyclic:$\measuredangle CBS=\measuredangle CTS=\measuredangle ITK$ $\measuredangle SBA=\measuredangle IBK$ $\measuredangle SBA=\measuredangle CBS\Rightarrow \measuredangle ITK=\measuredangle IBK$ Do this for $CSLI$ too. $\measuredangle SBA=\measuredangle IBK=\measuredangle ITK=\alpha$ $\measuredangle ACT=\measuredangle LCI=\measuredangle LSI=\beta$ $\measuredangle CBI+\measuredangle ICB+\measuredangle BIC=180\Rightarrow\measuredangle BIC=180-(\alpha+\beta)$ $\measuredangle BIC=\measuredangle BIT=\measuredangle BKT=\measuredangle AKD=180-(\alpha+\beta)\Rightarrow \measuredangle DKA=\alpha+\beta$ Do this to get $\measuredangle DLA=180-(\alpha+\beta)\Rightarrow \measuredangle ALD=\alpha+\beta$ too $\measuredangle AKD+\measuredangle KDA+\measuredangle DAK=180\Rightarrow\measuredangle ADK=\measuredangle ADL=\measuredangle IDK=\measuredangle IDL=90$ $ATBC$ cyclic:$\measuredangle ACT=\measuredangle ABT=\measuredangle KBT=\measuredangle KIT=\beta$ $ABCS$ cyclic:$\measuredangle SBA=\measuredangle SCA=\measuredangle SCL=\measuredangle SIL=\alpha$ $\measuredangle ITK+\measuredangle TKI+\measuredangle KIT=180\Rightarrow\measuredangle TKI=\measuredangle DKI=180-(\alpha+\beta)\Rightarrow \measuredangle IKD=\alpha+\beta$ $\measuredangle LSI+\measuredangle SIL+\measuredangle ILS=180\Rightarrow\measuredangle ILS=\measuredangle ILD=180-(\alpha+\beta)\Rightarrow\measuredangle DLI=\alpha+\beta$ $\measuredangle IKD+\measuredangle KDI+\measuredangle DIK=180\Rightarrow\measuredangle DIK=90-(\alpha+\beta)$ $\measuredangle DLI+\measuredangle LID+\measuredangle IDL=180\Rightarrow\measuredangle LID=90-(\alpha+\beta)$ Look at $AKIL$. We got $$\measuredangle IKL=\measuredangle IKD=\measuredangle DKA=\measuredangle LKA=\alpha+\beta$$$$\measuredangle ALK=\measuredangle ALD=\measuredangle DLI=\measuredangle KLI=\alpha+\beta$$$$\measuredangle KAI=\measuredangle KAD=\measuredangle DAL=\measuredangle IAL=90-(\alpha+\beta)$$$$\measuredangle LIA=\measuredangle LID=\measuredangle DIK=\measuredangle AIK=90-(\alpha+\beta)$$ $AKIL$ is a quadrilateral which its diagonals are angle bisectors,so $AKIL$ is a rhombus.

Sadigly wrote: Let $D$ be the intersection point of $AI$ and $KL$ $B;K;A$ colinear $C;L;A$ colinear $B;I;S$ colinear $C;I;T$ colinear $T;K;L;S;D$ colinear $\measuredangle CBS=\measuredangle SBA=\alpha$ $\measuredangle ACT=\measuredangle TCB=\beta$ $\measuredangle KAD=\measuredangle DAL=90-(\alpha+\beta)$ Claim:$BTKI$ is cyclic Proof: $BTSC$ cyclic:$\measuredangle CBS=\measuredangle CTS=\measuredangle ITK$ $\measuredangle SBA=\measuredangle IBK$ $\measuredangle SBA=\measuredangle CBS\Rightarrow \measuredangle ITK=\measuredangle IBK$ Do this for $CSLI$ too. $\measuredangle SBA=\measuredangle IBK=\measuredangle ITK=\alpha$ $\measuredangle ACT=\measuredangle LCI=\measuredangle LSI=\beta$ $\measuredangle CBI+\measuredangle ICB+\measuredangle BIC=180\Rightarrow\measuredangle BIC=180-(\alpha+\beta)$ $\measuredangle BIC=\measuredangle BIT=\measuredangle BKT=\measuredangle AKD=180-(\alpha+\beta)\Rightarrow \measuredangle DKA=\alpha+\beta$ Do this to get $\measuredangle DLA=180-(\alpha+\beta)\Rightarrow \measuredangle ALD=\alpha+\beta$ too $\measuredangle AKD+\measuredangle KDA+\measuredangle DAK=180\Rightarrow\measuredangle ADK=\measuredangle ADL=\measuredangle IDK=\measuredangle IDL=90$ $ATBC$ cyclic:$\measuredangle ACT=\measuredangle ABT=\measuredangle KBT=\measuredangle KIT=\beta$ $ABCS$ cyclic:$\measuredangle SBA=\measuredangle SCA=\measuredangle SCL=\measuredangle SIL=\alpha$ $\measuredangle ITK+\measuredangle TKI+\measuredangle KIT=180\Rightarrow\measuredangle TKI=\measuredangle DKI=180-(\alpha+\beta)\Rightarrow \measuredangle IKD=\alpha+\beta$ $\measuredangle LSI+\measuredangle SIL+\measuredangle ILS=180\Rightarrow\measuredangle ILS=\measuredangle ILD=180-(\alpha+\beta)\Rightarrow\measuredangle DLI=\alpha+\beta$ $\measuredangle IKD+\measuredangle KDI+\measuredangle DIK=180\Rightarrow\measuredangle DIK=90-(\alpha+\beta)$ $\measuredangle DLI+\measuredangle LID+\measuredangle IDL=180\Rightarrow\measuredangle LID=90-(\alpha+\beta)$ Look at $AKIL$. We got $$\measuredangle IKL=\measuredangle IKD=\measuredangle DKA=\measuredangle LKA=\alpha+\beta$$$$\measuredangle ALK=\measuredangle ALD=\measuredangle DLI=\measuredangle KLI=\alpha+\beta$$$$\measuredangle KAI=\measuredangle KAD=\measuredangle DAL=\measuredangle IAL=90-(\alpha+\beta)$$$$\measuredangle LIA=\measuredangle LID=\measuredangle DIK=\measuredangle AIK=90-(\alpha+\beta)$$ $AKIL$ is a quadrilateral which its diagonals are angle bisectors,so $AKIL$ is a rhombus. Yeah i did similar also here all angles that you say are equals then we see that ATI is equaliteral and it is easy to finish the prove

Easy::: $(ABC)\in \mathbb{S}^1$,$a=x^2,b=y^2,c=z^2$ so $s=-xz$ and $t=-xy$. $k=ST\cap AB$ And $l=ST \cap AC$ first prove that $a+j=k+l$ then $AI\bot KL$ then it is obvious that the figure that satisfies these conditions is indeed a rhombus.$\boxed{\lambda}$

From $SA=SI$ and $TA=TI$ we have that $ST$ is the perpendicular bisector of $AI$, so $AK=KI$, $AL=LI$ and $KL\perp AI$. Since $AI$ is also the angle bisector of $\angle KAL$, we conclude that $AI$ is the perpendicular bisector of $KL$, so $AK=AL$. Hence $KI=AK=AL=LI$, so $AKIL$ is a rhombus.