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.
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.