Lecture
Theorem
Any diametrical plane of the ball is its plane. The center of the ball is its center of symmetry.
Evidence
Let α be the diametrical plane and X be an arbitrary point of the ball. We construct the point X` symmetric to the point X with respect to the segment XX` and intersects with it in its middle. From the equality of right triangles OAX and OAX`, it follows that OX` = OX.
Since OX ≤ R, then OX` ≤ R, i.e. the point symmetric to the point X belongs to the ball. The first assertion of the theorem is proved.
Now let X`` be a point symmetric to X with respect to the center of the ball. Then OX`` = OX ≤ R, i.e. point X`` belongs to the ball. The theorem is proven completely.
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