I was going to make this into a problem, but I didn't know how to articulate it well enough for that purpose. Anyhow, it is kind of interesting. Looking at the diagram:
As you see, I've simply bisected the first circle, and put in another two circles half the diameter inside. Next, I use the vesica to bisect the original circle perpendicularly:
Now I draw segments from the centres of both smaller circles to the top point F:
Now, If we bisect these line segments, we get two points, H and I:
If we connect the two points, we get a line equal in distance to one-fourth of our first circle:
You will also notice that the line segments extended from bisecting the two original segments intersects at one-fourth the radius also.
We can prove this by drawing a vesica with the original circle, noting that it creates and angle of 60 degrees. Because of opposite interior angles, we can be sure that say angle ADF is also 60 degrees. Since we used perpendicular bisection, the top angle, AFD is 30 degrees. If we look at the 30, 60, 90 triangle:
The hypotenuse (for this example, that would be line segment DF) is 2x. Bisecting gives us x, which is the radius of our circle centred on point D or C. This is the measure of diameter, so half x is . Therefore, we have divided the original circle in quarters.
This is not really important, but maybe someone could make a problem out of it.