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Prove that G is a center of the circle circumscribed on BED.


So we know that we should prove that G lies on the perpendicular bisectors of every side of the BED triangle. For |ED| we can note that $\triangle EDG$ is equilateral so G, as the opposite vertex of |ED|, has to be on the perpendicular bisector.

And... I'm stuck. How can I show it for the other two? There is a ton of angles here but I don't know how to put them to use. Could you tell me what I should take into account while dealing with geometry problems like this? I don't know what to look at and the amount of information on the picture kind of makes me dizzy and everything looks useful and useless at the same time.

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And it's not enough and easier to show that $\overline{DG}=\overline{EG}=\overline{BG}$? – draks ... May 5 '12 at 17:26
It's undoubtly easier but is it enough? I mean - is there only one such point outside of every triangle that the distances to it from its vertices are equal and that point is the center of the circle in which that triangle is inscribed? – Straightfw May 5 '12 at 17:35
Oh my, what am I asking, of course it's enough.. :P OK, will try to break it now and will respond in a second! – Straightfw May 5 '12 at 17:41
Maybe you can show that the $\triangle GEB$ is equilateral... – draks ... May 5 '12 at 17:45
Thank you, I posted my answer using another thought below :) – Straightfw May 5 '12 at 18:01

Note that $EC=ED$ (isosceles triangle) and therefore $EC=EG$. So it is enough to show that $BG=EC$, for that will show that $G$ is equidistant from all three vertices of $\triangle EDB$.

Let $DC$ meet $EG$ at $X$, and let $DB$ meet $EG$ at $Y$. Join $CG$. Use angle-chasing to show that parallelograms $EBCY$ and $BCGX$ are congruent. Then $BG$ and $EC$ are corresponding diagonals of congruent parallelograms, so they are equal.

Remark: One can make the proof more attractive by extending $DG$ so that it meets $AC$ extended. Join $CG$ and $BG$. Now we have a very nicely symmetric picture, and a solution writes itself down.

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Thank you! Below, I posted my approach, if you didn't mind checking it. – Straightfw May 5 '12 at 18:02
@Straightfw: Looks as if we used same approach, and even used same names for intersection points! I am making a guess about what is brown and what is not, have some colour difficulty over the red brown green range. After a bit of waiting, you should accept your solution. – André Nicolas May 5 '12 at 18:11
Oh, that's right :) Thank you very much, will do as you say! – Straightfw May 6 '12 at 9:39
up vote 0 down vote accepted

I think I got it!

Let the brown line cross EG at X, red cross EG at Y. Then EDX is congruent to DGY (the same angles). Also, $\angle GED=60^\circ$ and, as $\angle DAC = 60^\circ$ as well, EG is parallel to AC. Let Z be a point on CD on the same level as F forming a line FZ parallel to EG and AC. Then BZ=CF (BDC is isosceles) and FE=ZG (altitudes of the congruent triangles). Hence, BZ+ZG=BG=EG=DG q.e.d.

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