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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
2 years, 1 month ago

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Angle bisector theorem and cevas

Xuming Liang - 2 years, 1 month ago

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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

Nihar Mahajan - 2 years, 1 month ago

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

Xuming Liang - 2 years, 1 month ago

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@Xuming Liang Lol , okay...

Nihar Mahajan - 2 years, 1 month 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 - 2 years, 1 month 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 - 2 years, 1 month ago

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

Saarthak Marathe - 2 years, 1 month ago

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Try proving it with another method.

Nihar Mahajan - 2 years, 1 month ago

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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/

Saarthak Marathe - 2 years, 1 month ago

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

Saarthak Marathe - 2 years, 1 month ago

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we need to prove (BD/DC)(CE/AE)(AF/BF) = 1

Dev Sharma - 2 years, 1 month ago

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

Nihar Mahajan - 2 years, 1 month ago

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my geometry is weak...please show proof.

Dev Sharma - 2 years, 1 month ago

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Which theorem?

Shivam Jadhav - 2 years, 1 month ago

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

Nihar Mahajan - 2 years, 1 month ago

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ceva theorm has something with concurrent....

Dev Sharma - 2 years, 1 month ago

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

Nihar Mahajan - 2 years, 1 month ago

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

Nihar Mahajan - 2 years, 1 month ago

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