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.

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TopNewestAngle bisector theorem and cevas

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Ah , I knew you would get it quickly. I used Ceva's instead of Menelaus. Nevermind Ceva's and Menelaus are brothers ;) :P

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I actually meant to type ceva...haven't used the two in a while so got them mixed up haha

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Suppose the circumcircle of \(DEF\) meets \(AD\) again at \(X\). Prove that the reflection of \(X\) over \(EF\) lies on \(BC\)

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Such an easy one. Use ceva's theorem and angle bisector property.

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Try proving it with another method.

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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/

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We can prove it by co-ordinate geometry but then it would be tedious.

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we need to prove

(BD/DC)(CE/AE)(AF/BF) = 1Log in to reply

Where's the proof? You only stated what to prove.

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my geometry is weak...please show proof.

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Which theorem?

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Now you can see that some people have already taken its name ;)

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ceva theorm has something with concurrent....

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It does not have "something" , it is fully based on concurrency :)

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I won't give away the name unless a solution is posted here ;)

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