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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, 4 months ago

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Angle bisector theorem and cevas

- 2 years, 4 months ago

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

- 2 years, 4 months ago

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

- 2 years, 4 months ago

Lol , okay...

- 2 years, 4 months ago

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 $$BC$$

- 2 years, 4 months ago

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 $$X' \in \overline{BC}$$

- 2 years, 4 months ago

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

- 2 years, 4 months ago

Try proving it with another method.

- 2 years, 4 months ago

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/

- 2 years, 4 months ago

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

- 2 years, 4 months ago

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

- 2 years, 4 months ago

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

- 2 years, 4 months ago

my geometry is weak...please show proof.

- 2 years, 4 months ago

Which theorem?

- 2 years, 4 months ago

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

- 2 years, 4 months ago

ceva theorm has something with concurrent....

- 2 years, 4 months ago

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

- 2 years, 4 months ago

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

- 2 years, 4 months ago