Given the regular heptagon, how to prove the four point in circle?? 
Given the regular heptagon, prove the four point in circle.


Here's my attempt to construct an isosceles trapezoid, but I do not know how to do next:

 A: Label points as in the picture.

Note that $\angle AGH = \dfrac{2\pi}{7} = \angle GHA$, hence $AH=AG$. Also, $\angle ICB = \dfrac{3\pi}{7} = \angle BIC$, therefore $BI=BC$. So $AH=AG=BC=BI$. Moreover $AE=BE$ and $\angle DBE = \dfrac{\pi}{7} = \angle DAE$. It follows from SAS that $\triangle HAE = \triangle IBE$. In particular $$\angle DHE = \pi - \angle EHA = \pi - \angle EIB = \angle DIE$$
from which it follows that $H, E, D, I$ are concyclic.
A: 
Let $a=\frac{2 \pi}{7}.$ Let us consider the heptagon $ABCDEFG$ with resp. coordinates $(\cos ka, \sin ka), k=0,1,\cdots 6$.
Let us denote by $W$ the center of the circumscribed circle to $(B,C,K)$. It is situated at the intersection of the two perpendicular bissectors of $BC$ and $BK$.
The perpendicular bissector of $BC$ passes through $O$ and $F$, due to the symmetry of the figure wrt line $FJ$. Therefore $$\vec{OW}=p\vec{OJ}$$ which is equivalent to the fact that, for a certain $p$:
$$(W_x,W_y)=p(\cos 3a/2,\sin 3a/2)\tag{1}$$
(because polar angle of $\vec{OJ}$ is $3a/2$).
Triangle $KBD$ is an homothetic image of $ADE$ because they have the same angles. It is therefore an isoceles triangle too with corresponding parallel bases $BK$ and $DE$ ; therefore parallel altitudes, altitude $DH$ being the perpendicular bissector of $BK$, we can conclude that $W$ and $D$ have the same ordinate $\sin(3a)$. Using (1), we can conclude that
$$(W_x,W_y)=2 \cos(3a/2)(\cos 3a/2,\sin 3a/2).\tag{2}$$
Let $H=AE \cap BF$. Due to the symmetry of the figure wrt line $GI$, we have, for a certain $q$::
$$\vec{OH}=q\vec{OG} \iff (H_x,H_y)=q(\cos(6a),\sin(6a))=q(\cos a, -\sin a)$$
The value of $q$ is easily found to be $q=2 \cos a -1$.
Now, if we turn to complex number representation setting $b:=a/2=\pi/7$, we have:
$$W=2 \cos(b)e^{ib}, \ \ H=(2 \cos(a)-1)e^{i5b}, \ \ B=e^{i2b}$$
it remains to check that the following squared distances are equal:
$$WB^2=WH^2$$
Meaning that:
$$|2\cos(b)e^{ib}-(2 \cos(2b)-1)e^{-2ib}|^2=|2\cos(b)e^{ib}-e^{2ib}|^2$$
which is true (verification using general trigonometry formulas and particular relationships specific to angle $a=\frac{2 \pi}{7}$ ; I can give more details).
Reamrk: a very nice article here on the heptagon.
