I need a simple proof that a line cannot intersect a circle at three distinct points.
Or a more geometric proof: If a circle intersects a line in $A$ and $B$, the center of the circle lies on the center normal of the line segment $AB$. If there is a third intersection point $C$, the center of the circle must also lie on the center normal of $BC$. But these two center normals are distinct parallel lines, and cannot have point in common.
Without loss of generality, assume the circle is $x^2 + y^2 = r^2$ and the line is $y = mx + c$.
The x coordinates of the point of intersection satisfy $x^2 + (mx+c)^2 = r^2$ which is a quadratic and hence has at most $2$ roots.
Since given an $x$, the $y$ on the line is uniquely determined, we are done.
Similar to Harald's proof, draw in a radius from the center of the circle to each point where the line intersects the circle. Now draw a perpendicular segment from the center to a point C on the line. Assuming we have more than one point of intersection, we have multiple right triangles which are congruent due to the HL theorem. Clearly we can't have a third point of intersection because there cannot be 3 distinct points along the line equidistant from C.
Take any 3 distinct points on a circle and notice that each angle of the triangle formed by those 3 points is higher than 0 and smaller than 180 degrees. Any of the angles formed by 3 distinct points on a line (degenerate triangle) takes a value of either 0 degrees or 180 degrees.
The proof is complete.