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.

## Comments

Sort by:

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

Log in to reply

– Nihar Mahajan · 1 year, 8 months ago

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

– Xuming Liang · 1 year, 8 months ago

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

– Nihar Mahajan · 1 year, 8 months ago

Lol , okay...Log in to reply

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

Log in to reply

– Nihar Mahajan · 1 year, 8 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 \(X' \in \overline{BC}\)Log in to reply

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

Log in to reply

– Nihar Mahajan · 1 year, 8 months ago

Try proving it with another method.Log in to reply

– Saarthak Marathe · 1 year, 8 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/Log in to reply

– Saarthak Marathe · 1 year, 8 months ago

We can prove it by co-ordinate geometry but then it would be tedious.Log in to reply

we need to prove

(BD/DC)– Dev Sharma · 1 year, 8 months ago(CE/AE)(AF/BF) = 1Log in to reply

– Nihar Mahajan · 1 year, 8 months ago

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

– Dev Sharma · 1 year, 8 months ago

my geometry is weak...please show proof.Log in to reply

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

Log in to reply

– Nihar Mahajan · 1 year, 8 months ago

Now you can see that some people have already taken its name ;)Log in to reply

– Dev Sharma · 1 year, 8 months ago

ceva theorm has something with concurrent....Log in to reply

– Nihar Mahajan · 1 year, 8 months ago

It does not have "something" , it is fully based on concurrency :)Log in to reply

– Nihar Mahajan · 1 year, 8 months ago

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