I'm trying to argue that any line in $\mathbb{R}^2$ that intersects the line segment $L$ along the $x$-axis from $- \frac{l}{2}$ to $\frac{l}{2}$, for some $l >0$, must make angle with the $x$ axis in the range $\theta \in [0, 2 \pi]$ and have distance from the origin, say $p$, in the range $[0, |\cos(\theta)|*l/2 ]$.
The first part is easy sin.c any line segment through the origin that makes angle $\theta \in [0, 2 \pi]$ with the $x$-axis will intersect $L.$
To show the range of $p,$ it seems to me that a vertical line at $x=l/2$ has distance $l/2$ from the origin and intersect $L$ with $\theta = \pi/2,$ while a vertical line at $x=0$ intersects $L$ and has distance $0$ from the origin. I am not sure why the max value for $p$ should be $l/2.$ Any insights appreciated.
The context of this question, for reference, is in an integral in Do Carmo's Differential Geometry of Curves and Surfaces at the bottom of page $44$ as part of an argument for the Cauchy-Crofton formula.
Best Answer
Proof (almost) without words.
The given line intersects the $x$-axis at a point $A.$ The blue and brown triangles are similar. Hence $$p\leq \frac{l}{2}\sin\theta,$$ where $p$ and $\frac{l}{2}\sin\theta$ are the lenghts of corresponding sides opposite to the angle $\theta.$