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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## Comments

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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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$\overline{BC} \cap \odot DEF = \{X'\}$ 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 $X' \in \overline{BC}$

That was nice. LetLog in to reply

Which theorem?

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

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

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

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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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