In this configuration, there is a triangle $ABC$ with 3 acute angles. Let there be a tangent line in $A$ to it's circumcircle, and let $P$ be the intersection of the tangent line and the line $BC$. Let $M$ be the middle of $PA$. Then let $R$ be the intersection of the circumcircle and $BM$. Then let $S$ be the intersection of the circumcircle and the line $PR$. Prove that $CS$ and $PA$ are parallel!
Proving two lines parallel
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geometry
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0Shows us what you tried to do. – 2017-01-10
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0@JohnMayne I've added it to the question ;) – 2017-01-10
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0@Joffan Yes, it has to be the mid-point, but I can't see what I can do what I can do with the information... – 2017-01-10
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0@Joffan Yes, sorry, it's true that I haven't been clear enough on that point... – 2017-01-10
1 Answers
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Note that $\angle RAM = \angle MBA$, so $\triangle RAM \sim \triangle MBA$, thus $$\frac{MR}{MA} = \frac{MA}{MB}.$$ Since $MA=MP$, the last equality yields $$\frac{MR}{MP}=\frac{MP}{MB}$$ so $\triangle MRP \sim \triangle MPB$ (by SAS). In particular $\angle MPR = \angle PBR$. Since $B,C,R,S$ are concyclic, we have $\angle CBR = \angle CSR$. Combining the last two equalities we get $\angle MPR = \angle CSR,$ so $AP \parallel CS$.
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0Very well done! +1! – 2017-01-10
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0Excellent, I probably would've started $\angle ARM=\angle MAB$ (if I had seen all this), but that doesn't take away from your solution. – 2017-01-10