Stuck on geometry proving Points S and D are respectively the center of the circle circumscribed on the acute triangle ABC and the orthocenter of this triangle. Prove that ASBX, where X is the center of the circle circumscribed on the triangle ABD, is a rhomb.
Could you please guide me where to even start? I'm really bad at geometry problems and when I came across ones like this, I don't even know what should I try to do and end up gazing at the picture and being unable to find anything useful...
Edit one day later: OK, I'm trying my best to beat it and here's what I have:
Let $\angle AXB=2\alpha$, then $\angle ADB = \alpha$. Similarly $\angle ASB=2\beta$ and $\angle ACB = \beta$. 
$|AS|=|SB|=R$ where R is the radius of the circle circumscribed on $\triangle ABC$. Then, $\triangle ASB$ is isosceless. Also $|AX|=|BX|=r$ where r is the radius of the circle circumscribed on $\triangle ADB$ so $\triangle ADB$ is isosceless as well.
And here I get stuck. We know that we have both isosceless triangles but I can't find a way to show that R=r which would prove all the sides being even. The "mutual" angle for both circles is $\angle ADB$ so I suppose I should use it. However, I don't think I can put it to use within the circle circumscribed on $\triangle ABC$ as its vertex doesn't lie on the circle. I thought about the law of sines but $\frac{|AB|}{\sin \alpha}=R$ and $\frac{|AB|}{\sin \beta}=r$ don't look promising. Could you give me a hint on how to move on?
Edit again: I believe I got it! Firstly, let's note something regarding what I wrote above and clarify: $\angle AXB=2\alpha$ so the reentrant angle $\angle AXB'=360^\circ-2\alpha$. Then, $\angle ADB=180^\circ-\alpha$.
Then, Let E be the foot of the altitude from B to $|AC|$ and let F be the foot of the altitude from A to $|BC|$. Let's have a look at AFDE quadrilateral. $\angle CEF=\angle CFD=90^\circ$. $\angle ECF = \angle ACB = \beta$ Then $\angle EDF=360^\circ - 90^\circ-90^\circ-\beta = 180^\circ-\beta$. $\angle EDF = \angle ADB$ so $\alpha = \beta$. Hence $\frac{|AB|}{\sin \alpha}=\frac{|AB|}{\sin \beta}=R=r$. All the sides are then equal and $|SX|$ is a perpendicular bisector of $|AB|$ so $ASBX$ is a rhomb q.e.d.
 A: So here's the answer I managed to get to with the help of Isaac:
Let $\angle AXB=2\alpha$ so the reentrant angle $\angle AXB'=360^\circ-2\alpha$. Then, $\angle ADB=180^\circ-\alpha$. Similarly $\angle ASB=2\beta$ and $\angle ACB = \beta$. 
$|AS|=|SB|=R$ where R is the radius of the circle circumscribed on $\triangle ABC$. Then, $\triangle ASB$ is isosceless. Also $|AX|=|BX|=r$ where r is the radius of the circle circumscribed on $\triangle ADB$ so $\triangle ADB$ is isosceless as well.
Let E be the foot of the altitude from B to $|AC|$ and let F be the foot of the altitude from A to $|BC|$. Let's have a look at CFDE quadrilateral. $\angle CEF=\angle CFD=90^\circ$. $\angle ECF = \angle ACB = \beta$ Then $\angle EDF=360^\circ - 90^\circ-90^\circ-\beta = 180^\circ-\beta$. $\angle EDF = \angle ADB$ so $\alpha = \beta$. Hence $\frac{|AB|}{\sin (180^\circ-\alpha)}=\frac{|AB|}{\sin \beta}=R=r$. All the sides are then equal so $ASBX$ is a rhomb q.e.d.
A: edit Up here are my revised hints; down at the bottom were my original hints, which I am less confident about.
As I stated in my comment, when I drew out the figure and thought about what it would mean for $ASBX$ to be a rhombus, it made me think that there should be more symmetry than was initially visible.  Specifically, it seemed like if I rotated $C$ by 180° about the midpoint of $\overline{AB}$ to $C'$, $X$ would be the circumcenter of $\triangle ABC'$.  So let's try to prove that fact.
Hint 1:

 With $C'$ drawn in, construct the circumcenter $S'$ and circumcircle of $\triangle ABC'$.  If this is the circumcenter of $ABD$, then $D$ lies on this circle.  So, show that $D$ lies on this circle.

Hint 2:

 In $\triangle ABC$, the altitudes from $A$ and $B$ intersect at $D$.  These altitudes are also perpendicular to $AC'$ and $BC'$.  Look at the angles in the quadrilateral $ADBC'$.

Having shown that $X$ is the circumcenter of $ABC'$...
Hint 3:

 ... $X=S'$ must be the rotation image of $S$ by 180° about the midpoint of $\overline{AB}$.  (Why?)

Hint 4:

 That $ASBX$ is a rhumbus follows directly.  (Why?)


I'd start by thinking about what characteristics of a rhombus can be used to prove that a quadrilateral is a rhombus and whether any of these characteristics relate to circumcenters, since two of the vertices of the rhombus are circumcenters of triangles.
Hint 1:

 The diagonals of a rhombus are perpendicular bisectors of each other.

Hint 2:

 $X$ and $S$ both lie on the perpendicular bisector of $\overline{AB}$.

Possible Hint 3:

 I'm pretty sure that the $ASBX$ is a rhombus even when the triangle is obtuse, so I'd guess that the restriction that the triangle is acute probably makes the proof easier somehow, perhaps by guaranteeing that $D$ is inside the triangle...

