With Hilbert's axiomatic system, How do I prove that a non-tangent line $d$ that intersects a circle $C$ intersects it in exactly two point?
My teacher gave us the following clue: First show that if ABC and A'B'C' are two triangles with right angles in B and B' and if AB≅A'B' and AC≅A'C' then the triangles are congruent.
On
Suppose $AC$ is the diameter of the circle, and $\ell$ is the line. Suppose that the point $B$ is an intersection of the line and the circle. We know that $ABC$ is a right triangle with a right angle at $B$. You want to find a point $B'$ such that $AB'C$ is a right triangle and such that $B'$ lies on $\ell$.
whats there to prove draw a circle draw a chord . the chord by definition joins any two points of a circle e g. diameter (longest chord) so you have proved by definition that secant(chord) cuts circle in exactly two points.