As shown in the figure: Let the center radius be 2. This is to calculate the coordinates of e, b, f and h, and then solve a circle equation according to the coordinates of BFH. If we can work it out, we can prove a three-point BFH*** circle, and then bring the coordinates of E into the same equation. If this relationship is met, what about the four-point EBFH and the four-point * * circle? So E(o, o)? The coordinate of b (root number 2, root number 2)F(- root number 2, root number 3 2) f can be obtained by finding the intersection point p between the Y axis and BF, and then finding a straight line according to P and B, as shown in the following figure: b (root number 2, root number 2)? P (0,2 root number 2) Find the linear equation according to the two-point formula: (x-x1)/(x2-x1) = (y-y1)/(y2-y1): root number 2x+ root number 2y.

Let the equation of EBFH circle be X 2+Y 2+DX+EY+F = 0. Bring the coordinates of BFH point into the equation X 2+Y 2+DX+EY+F = 0?

Find D= root number 2? E=-3 root 2F=0, so the equation of the circle of EBHF is x2+y2+ root 2x-3 root 2y=0.

Then the equation of point E (0,0) is brought into the equation satisfying the circle. So EBFH is round at four o'clock!