TelvCohl wrote: Let $ A, B, D, E, F, C $ be six points lie on a circle (in order) satisfy $ AB=AC $ . Let $ P=AD \cap BE, R=AF \cap CE, Q=BF \cap CD, S=AD \cap BF, T=AF \cap CD $ . Let $ K $ be a point lie on $ ST $ satisfy $ \angle QKS=\angle ECA $ . Prove that $ \frac{SK}{KT}=\frac{PQ}{QR} $ Briefly solution According to Pascal's theorem we have $ST || BC$ so $ QK || AE$. + $\triangle QKS \sim \triangle RCA ; \triangle QKT \sim PBA$ +) $\frac{QK}{KS} = \frac{RC}{CA}$ and $\frac{QK}{KT} = \frac{PB}{AB} \rightarrow \frac{KS}{KT} = \frac{PB}{RC}$ + We need to prove $ \frac{PQ}{PB} = \frac{RQ}{RC}$ $\leftrightarrow \frac{PQ}{AP}. \frac{AP}{PB} = \frac{RQ}{AR}. \frac{AR}{RC}$ $\leftrightarrow \frac{sin \angle PAQ}{sin \angle PAB}= \frac{sin \angle RAQ}{sin \angle RAC}$ $\leftrightarrow \frac{MD}{DB} = \frac{MF}{FC}$ with $M$ is the intersection of $AQ$ and the circle. $\leftrightarrow \frac{MD}{MF} = \frac{DB}{FC}$ (*) +) Applying Cevasin theorem in triangle $ADF$ we have (*).

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My solution: From Pascal theorem ( for $ AADCBF $ ) we get $ BC \parallel ST $ , so from Reim theorem we get $ D, F, S, T $ are concyclic . From Pascal theorem ( for $ ADCEBF $ ) we get $ P, Q, R $ are collinear . Since $ \triangle ACR \sim \triangle SKQ $ and $ \triangle ABS \sim \triangle TKQ $ , so we get $ \frac{SK}{KT}=\frac{BP}{CR} $ . ... $ (\star) $ Since $ \frac{BP}{PQ}=\frac{\sin \angle RQF}{\sin \angle FCR}=\frac{CR}{QR} $ ( $ \because FA $ bisect $ \angle BFC $ ) , so combine with $ (\star) $ we get $ \frac{SK}{KT}=\frac{BP}{CR}=\frac{PQ}{QR} $ . Q.E.D

Keeping in mind that $AB = AC$, it follows form Pascal's Theorem on $CDAABF$ that $BC \parallel ST.$ Meanwhile, from Pascal's Theorem on $ADCEBF$, it follows that $P, Q, R$ are collinear. Then note that $\angle QKS = \angle ECA = \tfrac{1}{2}\widehat{AE}$ and $\angle QSK = \angle FBC = \tfrac{1}{2}\widehat{CF}.$ Therefore, $\angle SQK = \tfrac{1}{2}\left(\widehat{AC} + \widehat{EF}\right) = \angle ARC.$ Similarly, $\angle TQK = \angle APB.$ Now, from the Law of Sines applied to $\triangle QSP$ and $\triangle QTR$, we obtain \[\frac{QP}{QS} \cdot \frac{QT}{QR} = \frac{\sin\angle QSP}{\sin\angle QPS} \cdot \frac{\sin\angle QRT}{\sin\angle QTR}.\] Observe that $\angle QSP = \tfrac{1}{2}\left(\widehat{AB} + \widehat{DF}\right) = \tfrac{1}{2}\left(\widehat{AC} + \widehat{DF}\right) = \angle QTR.$ Therefore, we have \[\frac{QP}{QS} \cdot \frac{QT}{QR} = \frac{\sin\angle QRT}{\sin\angle QPS} \implies \frac{QP}{QR} \cdot \frac{QT}{QS} = \frac{\sin\angle ARP}{\sin\angle APR} = \frac{AP}{AR}.\] Now, note that $\angle ABP = \angle ABE = 180^{\circ} - \angle ACE = 180^{\circ} - \angle ACR.$ Therefore, it follows from the Law of Sines applied to $\triangle ABP$ and $\triangle ACR$ that \[\frac{AP}{AB} \cdot \frac{AC}{AR} = \frac{\sin\angle ABP}{\sin\angle APB} \cdot \frac{\sin\angle ARC}{\sin\angle ACR} \implies \frac{AP}{AR} = \frac{\sin\angle ARC}{\sin\angle APB}.\] Finally, by the Ratio Lemma in $\triangle QST$, we obtain \[\frac{QP}{QR} \cdot \frac{QT}{QS} = \frac{\sin\angle ARC}{\sin\angle APB} = \frac{\sin\angle SQK}{\sin\angle TQK} = \frac{KS}{KT} \cdot \frac{QT}{QS}.\] Simplification yields $\tfrac{QP}{QR} = \tfrac{KS}{KT}$ as desired. $\square$

My solution:(similar TelvCohl's solution) From Pascal theorem for $(DAFBEC)$ and $(FAADCB)$ we find $P,Q,R$ collinear and $ST\parallel BC.$ Also we know $\angle TDS=\angle CDA=\angle CBA=\angle ACB=\angle TFS\to DSTF$ is cyclic. Also we can find easily $\triangle ABP\sim \triangle TKQ,(\angle ABP=\angle TKQ,\angle PAB=\angle QTK),$ and $\triangle ACR\sim \triangle SKQ,(\angle ACR=\angle SKQ,\angle RAC=\angle QSK).$ Then we find this similarity $$\frac{SK}{KT}=\frac{BP}{CR}=^{?}\frac{PQ}{QR}.$$In $\triangle BPQ,\triangle RQF,\triangle CRF$ we apply The Sine Law we can find easily.

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Notice that $\angle SDT=\angle ADC=\angle AFB=\angle SFT$ hence $S,T,F,D$ are concyclic, so $ST\|BC$ by Reim's theorem on $(STFD)$ and $(BCDF)$. $\blacksquare$ CLAIM 2. Let $KQ$ meet $BC$ at $Y$, let $AE$ meet $BC$ at $X$, then $AE\|YQ$. Proof. By shooting lemma $\triangle AXC\sim \triangle ACE$ hence $$\angle BYQ=\angle SKQ=\angle ACE=\angle YXE$$$\blacksquare$ Therefore, $$\frac{SK}{KT}=\frac{SQ}{QT}\cdot\frac{\sin\angle SQK}{\sin\angle TQK}\qquad(1)$$Meanwhile $P,Q,R$ are collinear by Pascal's theorem, hence $$\frac{PQ}{QR}=\frac{PQ}{\sin\angle PSQ}\cdot\frac{QR}{\sin\angle QTR}=\frac{SQ}{QT}\frac{\sin\angle ARP}{\sin\angle APR}\qquad(2)$$Now we have $$\angle SQK=\angle (AE,BF)=\widehat{AB}+\widehat{EF}=\widehat{AC}+\widehat{EF}=\angle ERF\qquad(3)$$and similarly $$\angle TQK=\angle DPE\qquad(4)$$Since $\angle PEA=\angle PER$, by Ceva's theorem on $\triangle EPR$ and point $A$ we have $$\frac{\sin\angle ERF}{\sin\angle ARP}\cdot\frac{\sin\angle APR}{\sin\angle DPE}=1\qquad(5)$$Combining $(1),(2),(3),(4),(5)$ yields the result.