Butterfly Theorem Let $ABC$ an isosceles acute triangle and $D$ be the middle of the base $AB$.
Let's consider $E \in AB$ and $O$, the circumcenter of the triangle $ACE$.
Prove that the perpendicular in $D$ on $OD$, the perpendicular from $E$ to $BC$, and the parallel through $B$ at $AC$ are concurrent.
I tried to apply Butterfly Theorem from here.

 A: I don't know how Butterfly Theorem should be applied here but there is an algebraic solution to this problem (here is another solution to another problem with the same approach). WLOG, let's place your figure on the Cartesian coordinate system such that:
$$A=(-1,0),\\ B=(1,0), \\ C=(0,a), \\ E=(b,0), \\ D=(0,0).$$
First, we need to calculate the coordinates of $O$, which is the intersection of the lines $x=\frac{-1+b}{2}$ (perpendicular bisector of $AE$) and $y=\frac{-1}{a}x+\frac{a}{2}-\frac{1}{2a}$ (perpendicular bisector of $AC$). So, we have:
$$O=(\frac{-1+b}{2}, \frac{a^2-b}{2a}).$$

Now, assume the perpendicular to $OD$ at $D$ and the perpendicular to $BC$ from $E$ intersect each other at $M$.
It is very easy to see that the equation of $DM$ is:
$$y=\frac{a-ab}{a^2-b}x,$$
and the equation of $EM$ is:
$$y=\frac{1}{a}x-\frac{b}{a}.$$
Thus,
$$M=(\frac{a^2-b}{a^2-1}, \frac{a-ab}{a^2-1}).$$
To prove the claim, we only need to show that the slope of the line passing through $M$ and $B$ is equal to the slope of the line $AC$, which is $a$.
To do so,
$$\frac{\frac{a-ab}{a^2-1}-0}{\frac{a^2-b}{a^2-1}-1}= \frac{a-ab}{a^2-b-a^2+1}=a.$$
We are done.
A: Geometric solution:I could not construct an isosceles triangle with this statement, it became an equilateral triangle. May be it is a particular case. Draw like: A is top vertex, clockwise marking B is right and C is left of the base.
Find the mirror of E about BC, mark it as F.Mark intersection of EF with BC as H. connect O to F. Draw a circle on diameter OF, clearly it passes point D. EF is perpendicular on BC, and:
$\triangle EHB=\triangle FHB\Rightarrow \angle CBE=\angle CBF$
But: $\angle CBE=\angle ACB\Rightarrow \angle ACB=\angle CBF$
This means that $BF||AC$.
Now we have to show B is on the circle with diameter OF. This is clear because BO is altitude of triangle ABC and is perpendicular on AC and thereby on BF, which means $\angle OBF = 90^o$ which in turn means B is on the circle with diameter OF.
