Prove that a line that intersect with a convex plane curve >=3 times, it is tangent to the curve.

Question

I saw this lemma in Diff' geometry class. The proof (not formal) from the recitation is: say L intersects with a convex plane curve in a, b and c so that b is in [a,c]. Let L' be the tangent line of the curve in point b. The curve is convex so a and c are on the same side of L', hence L'=L and L is tangent to the curve......???? I don't understand how this prove(or show) anything and more importantly: How can a tangent line of a convex curve cross it in another 2 or more point?? By definition a closed simple convex curve creates a convex subset of R^2.

Answer 1

Suppose two distinct lines $L$ and $L'$ pass through $b$ and $a,c\in L$. Then if $b$ is between $a$ and $c$, it follows that $a$ and $c$ lie on opposite sides of $L'$. If we assume $C$ is convex and $L'$ is tangent to $C$ at $b$, then $a$ and $c$ must lie on the same side of $L'$. Thus, $L$ and $L'$ cannot be distinct and must coincide. (The only way this can happen, then, is for $L$ to be tangent to $C$ all along $[a,c]$, so the curve $C$ contains the line segment $[a,c]$. The curve is then not strictly convex, but is still, according to some definitions, convex.)