Proof of circle bisecting method When I was 7 or 8 years old, I was playing with drawing circles. For some reason, I thought about taking a 90° angle and placing the vertex on the circle and marking where the two sides intersected the circle.

It appeared to me that connecting the two points of intersection created a line through the center regardless of how the square was rotated.

I tested this many times and, within the error my tools, my conjecture seemed to be correct. I had indeed discovered a method for finding the center of a circle. Simply create at least two of these bisecting lines and voila!

The problem I have now, many years later, is that I am not satisfied with simply checking a bunch of times; I want a proof.
I can intuitively explain two extremes and the middle without any fancy math, but I still have not come up with a general proof.
As one of the points of intersection approaches the vertex of the angle, the leg with that point approaches a tangent of the circle. It is intuitive to me that a line, which is perpendicular to a tangent line and passes through the point at which the tangent line touches the circle, would bisect the circle.

In addition, when the angle is rotated such that the two line segments formed between the vertex and the points of intersection are equal, these line segments form half of an inscribed square. A line that passes through two opposite vertices of that square would also bisect the circle.

Can you prove whether placing the vertex of a 90° angle on a circle creates a bisector by connecting the points of intersection?
 A: This is (essentially) the converse of Thales' theorem:

The converse of Thales' theorem is then: the center of the circumcircle of a right triangle lies on its hypotenuse. (Equivalently, a right triangle's hypotenuse is a diameter of its circumcircle.)

The above quotation comes from the wikipage for Thales' Theorem, and the linked section contains three different proofs of this fact.
A: Suppose that a triangle like the one in the picture has the median $AM$ equal to the half side $CB$ that is $CM=MB$.
You have an isosceles triangle $MCA$ with two equal angles and another, different, isosceles triangle $MBA$. 
Now let's sum the angles $x+x+y+y=2x+2y=2(x+y)$ must give $180°$ which means that 
$2(x+y)=180°$ that is $x+y=90°$.
Thus we have proved that if the midpoint of the longest side of a triangle has the same distance from the vertices the triangle is rectangle and we can draw a circle with centre $M$ and radius $MA$, the circumscribed circle. 

Now let's assume that the triangle is rectangle and draw from the vertices of the longest side the parallels to the other shorter sides. They intersect at $D$ and is easy to see that $ABCD$ is  a rectangle, whose diagonals are equal and cut each other in their midpoint. So it follows that $AM=BM=CM$.
We have proved that in a triangle rectangle the midpoint of the longest side has the same distance from the three vertex and is the centre of the circumscribed circle, that is the reason why when you where a child the right angles had sides which intersect the circle in points that are the diameter of the circle.
Hope this helps

