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Simple concurrency

In \(\Delta ABC\) , let \(D\) be any point on \(\overline{BC}\).Let \(\overline{DF} \ , \ \overline{DE}\) be bisectors of \(\angle ADB \ , \ \angle ADC\) respectively with \(E \in \overline{AC} \ , \ F \in \overline{AB}\) . Prove that \(\overline{AD} \ , \ \overline{BE} \ , \ \overline{CF}\) are concurrent.

Posed this while playing with a theorem ;)

Note by Nihar Mahajan
1 year, 8 months ago

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Angle bisector theorem and cevas Xuming Liang · 1 year, 8 months ago

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@Xuming Liang Ah , I knew you would get it quickly. I used Ceva's instead of Menelaus. Nevermind Ceva's and Menelaus are brothers ;) :P Nihar Mahajan · 1 year, 8 months ago

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@Nihar Mahajan I actually meant to type ceva...haven't used the two in a while so got them mixed up haha Xuming Liang · 1 year, 8 months ago

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@Xuming Liang Lol , okay... Nihar Mahajan · 1 year, 8 months ago

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@Nihar Mahajan Can you prove the following simple fact (I tried to devise a problem out of this configuration):

Suppose the circumcircle of \(DEF\) meets \(AD\) again at \(X\). Prove that the reflection of \(X\) over \(EF\) lies on \(BC\) Xuming Liang · 1 year, 8 months ago

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@Xuming Liang That was nice. Let \(\overline{BC} \cap \odot DEF = \{X'\}\) and I claim that \(X'\) is the reflection of \(X\). To prove this its easy to see that \(\Delta XFE \cong \Delta X'FE\) which makes \(\Box XFX'E\) a kite. This implies \(\overline{XX'} \perp \overline{FE}\) and \(XK=KE\) provided that \(\overline{FE} \cap \overline{XX'} = \{K\}\). Thus it is proved that \(X'\) is the reflection of \(X\) such that \(X' \in \overline{BC}\) Nihar Mahajan · 1 year, 8 months ago

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Such an easy one. Use ceva's theorem and angle bisector property. Saarthak Marathe · 1 year, 8 months ago

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@Saarthak Marathe Try proving it with another method. Nihar Mahajan · 1 year, 8 months ago

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@Nihar Mahajan Nihar, I suppose u are an Indian. If u like geometry try solving INMO 1994 question number 5. It is in one of my notes(rather the only note). https://brilliant.org/discussions/thread/amazing-geometry-question/ Saarthak Marathe · 1 year, 8 months ago

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@Nihar Mahajan We can prove it by co-ordinate geometry but then it would be tedious. Saarthak Marathe · 1 year, 8 months ago

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we need to prove (BD/DC)(CE/AE)(AF/BF) = 1 Dev Sharma · 1 year, 8 months ago

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@Dev Sharma Where's the proof? You only stated what to prove. Nihar Mahajan · 1 year, 8 months ago

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@Nihar Mahajan my geometry is weak...please show proof. Dev Sharma · 1 year, 8 months ago

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Which theorem? Shivam Jadhav · 1 year, 8 months ago

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@Shivam Jadhav Now you can see that some people have already taken its name ;) Nihar Mahajan · 1 year, 8 months ago

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@Shivam Jadhav ceva theorm has something with concurrent.... Dev Sharma · 1 year, 8 months ago

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@Dev Sharma It does not have "something" , it is fully based on concurrency :) Nihar Mahajan · 1 year, 8 months ago

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@Shivam Jadhav I won't give away the name unless a solution is posted here ;) Nihar Mahajan · 1 year, 8 months ago

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