The diagonals $AD$, $BE$, $CF$ of cyclic hexagon $ABCDEF$ intersect in $S$ and $AB$ is parallel to $CF$ and lines $DE$ and $CF$ intersect each other in $M$. Let $N$ be a point such that $M$ is the midpoint of $SN$. Prove that circumcircle of $\triangle ADN$ is passing through midpoint of segment $CF$.
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Tags: geometry
TelvCohl
25.05.2015 21:18
My solution : Let $ T $ be the midpoint of $ CF $ . From Reim theorem ( for B-S-E and A-S-D ) $ \Longrightarrow MS $ is tangent to $ \odot (SDE) $ at $ S $ , so $ {MS}^2=MD \cdot ME=MC \cdot MF \Longrightarrow (C,F;S,N)=-1 \Longrightarrow SC \cdot SF=ST \cdot SN $ , hence we get $ SA \cdot SD= ST \cdot SN $ $ \Longrightarrow $ the midpoint $ T $ of $ CF $ lie on the circumcircle of $ \triangle ADN $ . Q.E.D
coldheart361
05.10.2019 11:14
TelvCohl wrote: My solution : Let $ T $ be the midpoint of $ CF $ . From Reim theorem ( for B-S-E and A-S-D ) $ \Longrightarrow MS $ is tangent to $ \odot (SDE) $ at $ S $ , so $ {MS}^2=MD \cdot ME=MC \cdot MF \Longrightarrow (C,F;S,N)=-1 \Longrightarrow SC \cdot SF=ST \cdot SN $ , hence we get $ SA \cdot SD= ST \cdot SN $ $ \Longrightarrow $ the midpoint $ T $ of $ CF $ lie on the circumcircle of $ \triangle ADN $ . Q.E.D
coldheart361
05.10.2019 11:16
TelvCohl wrote: My solution : Let $ T $ be the midpoint of $ CF $ . From Reim theorem ( for B-S-E and A-S-D ) $ \Longrightarrow MS $ is tangent to $ \odot (SDE) $ at $ S $ , so $ {MS}^2=MD \cdot ME=MC \cdot MF \Longrightarrow (C,F;S,N)=-1 \Longrightarrow SC \cdot SF=ST \cdot SN $ , hence we get $ SA \cdot SD= ST \cdot SN $ $ \Longrightarrow $ the midpoint $ T $ of $ CF $ lie on the circumcircle of $ \triangle ADN $ . Q.E.D Pardon, I am a beginner, could you explain what happens after the equation MD.ME=MC.MF many thanks
LKira
05.10.2019 11:34
coldheart361 wrote: TelvCohl wrote: My solution : Let $ T $ be the midpoint of $ CF $ . From Reim theorem ( for B-S-E and A-S-D ) $ \Longrightarrow MS $ is tangent to $ \odot (SDE) $ at $ S $ , so $ {MS}^2=MD \cdot ME=MC \cdot MF \Longrightarrow (C,F;S,N)=-1 \Longrightarrow SC \cdot SF=ST \cdot SN $ , hence we get $ SA \cdot SD= ST \cdot SN $ $ \Longrightarrow $ the midpoint $ T $ of $ CF $ lie on the circumcircle of $ \triangle ADN $ . Q.E.D Pardon, I am a beginner, could you explain what happens after the equation MD.ME=MC.MF many thanks $(C,F;S,N)$ is called cross ratio or double ratio
coldheart361
05.10.2019 11:39
LKira wrote: coldheart361 wrote: TelvCohl wrote: My solution : Let $ T $ be the midpoint of $ CF $ . From Reim theorem ( for B-S-E and A-S-D ) $ \Longrightarrow MS $ is tangent to $ \odot (SDE) $ at $ S $ , so $ {MS}^2=MD \cdot ME=MC \cdot MF \Longrightarrow (C,F;S,N)=-1 \Longrightarrow SC \cdot SF=ST \cdot SN $ , hence we get $ SA \cdot SD= ST \cdot SN $ $ \Longrightarrow $ the midpoint $ T $ of $ CF $ lie on the circumcircle of $ \triangle ADN $ . Q.E.D Pardon, I am a beginner, could you explain what happens after the equation MD.ME=MC.MF many thanks $(C,F;S,N)$ is called cross ratio or double ratio Thanks