# 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.

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.

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...

• Thank you. How do you know that |XS| is the perpendicular bisector of |AB|, though? – Straightfw May 2 '12 at 23:39
• @Straightfw: $S$ is the circumcenter of $\triangle ABC$, so it lies on the perpendicular bisectors of each side of the triangle, including $\overline{AB}$; and $X$ is the circumcenter of $\triangle ABD$, so it lies on the perpendicular bisectors of each side of that triangle, including $\overline{AB}$; so both $S$ and $X$ lie on the perpendicular bisector of $\overline{AB}$. – Isaac May 2 '12 at 23:44
• Wow, that's absolutely right! Thank you very much. Could you take a look at the initial question and give me some feedback? I tried to crack the proving of all the sides being even. – Straightfw May 2 '12 at 23:51
• @Straightfw: Looking at your edit and based on my sketches, I believe that $X$ and $S$ are on opposite sides of $\overline{AB}$, so either $2\alpha$ or $2\beta$ should be $2\pi-2\alpha$ or $2\pi-2\beta$... but I think I may have sent you down a messier-than-necessary path. What I've been pondering since yesterday is this: if $ASBX$ is a rhombus, that implies a certain amount of symmetry in the problem. In particular, if you rotate $C$ by 180° about the midpoint of $\overline{AB}$ to $C'$, it looks like $X$ should be the circumcenter of $\triangle ABC'$... I'll post new hints in a few mins. – Isaac May 3 '12 at 0:10
• @Straightfw: Except for a few very minor things, it looks good to me. (1) "Let's have a look at AFDE quadrilateral" should probably be CFDE; (2) $\frac{|AB|}{\sin \alpha}$ should probably be $\frac{|AB|}{\sin(180°-\alpha)}$ (which has the same value, but the angle in the Law of Sines is $180°-\alpha$ in this revision); and (3) all the sides equal $\implies$ rhombus without needing the perpendicular bisector at all. I'd suggest you put your solution into an answer and accept that answer. – Isaac May 3 '12 at 0:53