Let $M$ be the midpoint of $BC$. Let the tangents at $B$ and $C$ intersect at $D$. We make the following claim to remove the points $A$, $E$, $F$, and $K$. Claim: $DLMR$ is concyclic Elementary angle-chasing gives that $CMLF$ and $BEKM$ are both concyclic. $$\angle DRL=\frac{1}{2}\angle DKL=\frac{1}{2}\angle EKB=\frac{1}{2}\angle EMB=\angle ACB$$$$\angle DML=\angle MLC-\angle MDC=\angle MFC-\angle MDC=\angle FCB-\angle MDC=\angle ACB$$ Let $O$ be the circumcenter of $ABC$, let $(DLMR)$ intersect $BC$ again at $G$, and let $N$ be the midpoint of $RS$. Notice that $\angle GRD=\angle GLD=90^{\circ}$. It is sufficient to show that $NO=RN+OB$. Let $DB=DC=1$, $\theta = \angle DBC$, $\cos(\theta)=a$, and $\sin(\theta)=b$. Then, $$NO=RN+OB\iff NO^2=BN^2+BO^2=RN^2+BO^2+2 BO\cdot RN\iff BN^2=RN^2+2 BO\cdot RN\iff$$$$(2BN)^2=(2RN)^2+8BO\cdot RN\iff (BR+BS)^2=(BR-BS)^2+4BO\cdot RS \iff BR\cdot BS=BO\cdot RS$$Since $BO=\cot(\theta)$ it is sufficient to show that $\frac{BS-BR}{BS\cdot BR}=\tan(\theta)=\frac{b}{a}$. Standard but non trivial computation show that $$BR=2ab$$$$BS=\frac{2ab}{a^2-b^2}$$From here the rest is computation.

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Hopefully there exists an elegant synthetic solution. unrelated 12 problems=perfect distribution. CGA ANC GAN NCG

Shayan-TayefehIR wrote: For a triangle $\triangle ABC$ with an obtuse angle $\angle A$ , let $E , F$ be feet of altitudes from $B , C$ on sides $AC , AB$ respectively. The tangents from $B , C$ to circumcircle of triangle $\triangle ABC$ intersect line $EF$ at points $K , L$ respectively and we know that $\angle CLB=135$. Point $R$ lies on segment $BK$ in such a way that $KR=KL$ and let $S$ be a point on line $BK$ such that $K$ is between $B , S$ and $\angle BLS=135$. Prove that the circle with diameter $RS$ is tangent to circumcircle of triangle $\triangle ABC$. Proposed by Mehran Talaei Let M be the midpoint of BC, J be the intersection of two tangents from B and C. Through M, right angle BC; intersects BF at I. By simple angle chasing we have I L F C M are concyclic, and S L I are collinear. We have JLM = 180o - MLC = 180o - MFC = 180o - MCF = MLF = MRB so J L M R are concyclic. Let LI intersects BC at N, then JLN = JMN = 90o so J L M N R are concyclic. Let (CNL) intersects (O) again at T, then TLN = TCN = TBS, so S L T B are concyclic. => BTS = BLS = 135o. Let L' be the reflection of L through BC. Beacuse CTN = 90o, then CTNL' are concyclic. Because MR = ML, MN is the external bisector of angle RNL => R N L' are collinear, so CL' // BR => RBL' = 45o. We will prove that BRTL' are concyclic. Let LL' intersects BC at H. => BRHL' are concyclic. Simple angle chasing: TBH = TCL = TL'L = TL'H so BRHL' concyclic => BRTL' are concyclic. => RTL = 135o. => RTS = BTL' = 90o => And we can simply finish the problem by drawing the tangent of (RS).

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