# Condition for three lines to be concurrent.

A triangle $\bigtriangleup ABC$ is given, and let the external angle bisector of the angle $\angle A$ intersect the lines perpendicular to $BC$ and passing through $B$ and $C$ at the points $D$ and $E$, respectively. Prove that the line segments $BE$, $CD$, $AO$ are concurrent, where $O$ is the circumcenter of $\bigtriangleup ABC$.

I've seen this problem on AOPS site,and I've read that the condition for $BE,CD,AO$ to be concurrent is the following :

$$\frac{\sin BAO}{\sin OAC}.\frac{\sin ACD}{\sin DCB}.\frac{\sin EBC}{\sin EBA}=1$$

But I can't see the reason behind it... I know both Ceva and Menelaus's Theorem but I don't see how one of them is applied to give the condition above.

So I am asking if there's some theorem I am missing out or if that condition is just a rearranged form of Ceva\Menelaus's Theorem applied to some triangle I can't see.

• That is just the trigonometric form of Ceva's theorem. Commented Nov 11, 2015 at 13:55
• @ Jack D'Aurizio That isn't helping really much..What triangle do I have to consider ? Commented Nov 11, 2015 at 14:35
• I think Hoseyn Heydari's answer is pretty clear. Commented Nov 11, 2015 at 14:36
• I've just realized that my diagram was completely wrong according to the problem architecture so that's why I wasn't able to see why that was just Ceva's Theorem(because clearly the above condition doesn't apply to the triangle in the above diagram),in fact I've considerered the normal bisector of $\angle A$ and for some "coincidence" it happens that lines $BE,CD,AO$ are still concurrent.Do you know why ? Commented Nov 11, 2015 at 16:15
• From Galois theory ("extraversion" in Conway's language), it is clear that if the claim is true for the external angle bisector, then it must be equally true for the internal angle bisector (since all the other constructions involved are rational, so the external and internal angle bisectors are Galois conjugates). So the picture isn't wrong -- it just shows an equivalent problem. This is assuming that "line segments" means "lines"; if you really mean line segments, then the problem has a whole new level of difficulty added to it. Commented Mar 5, 2019 at 18:32

Lemma: In triangle $ABC$ with $D$ on $BC$ we have $\frac{BD}{CD}=\frac{AB}{AC}\times\frac{\sin(BAD)}{\sin(CAD)}$
Proof: We write Sin Theorem in triangles $ABD$ & $ACD$ then we have: $$\frac{\sin(BAD)}{BD}=\frac{\sin(ADB)}{AB}\quad \frac{\sin(CAD)}{CD}=\frac{\sin(ADC)}{AC}$$ since $$\sin(ADC)=\sin(\pi-ADC)=\sin(ADB)$$ we have: $$\frac{\sin(BAD)\times AB}{BD} = \frac{\sin(CAD)\times AC}{CD} .\square$$ By the lemma above we have if the lines meet segments $BC,CA,AB$ respectively at $D,E,F$ then: $$\frac{BD}{DC}\times\frac{CE}{EA}\times\frac{AF}{FB}=\frac{AB}{AC}\times\frac{\sin(BAD)}{\sin(CAD)}\times\frac{BC}{BA}\times\frac{\sin(CBE)}{\sin(ABE)}\times\frac{AC}{BC}\times\frac{\sin(ACF)}{\sin(BCF)}$$ So with canceling segments we have Sin Form of Ceva theorem: $$\frac{BD}{DC}\times\frac{CE}{EA}\times\frac{AF}{FB}=1\iff \frac{\sin(BAD)}{\sin(CAD)}\times\frac{\sin(CBE)}{\sin(ABE)}\times\frac{\sin(ACF)}{\sin(BCF)} =1$$ which is you can see.