I'm reading Differential Geometry by Heinrich W. Guggenheimer and I have a doubt about the proof of Schur's Theorem for Convex Plane Curves on page 31. I will put the theorem and the proof here before I say what are my doubts.
$\textbf{Theorem 2-19.}$ Given two curves $f, g$ with continuous tangents and piecewise continuous curvature, both of length $L$. If $|k_f(s)| \geq |k_g(s)|$, $k_f(s) \neq k_g(s)$, the chord subtended by $g$ is bigger than that subtended by $f$.
$\textbf{Proof.}$
We place both arcs in the lower half plane $x_2 \leq 0$ with endpoints on the $x_1$ axis, $x_{2f}(0) = x_{2f}(L) = x_{2g}(0) = x_{2g}(L) = 0$ and such that $x_{1f} (L) > x_{1f} (0)$, $x_{1g} (L) > x_{1g} (0)$. In this case, both curvatures are non-negative. The chords subtended are $d_f = x_{1f}(L) - x_{1f}(0) = \int_0^L \cos \theta_f (s) ds$, $d_g = x_{1g}(L) - x_{1g}(0) = \int_0^L \cos \theta_g (s) ds$
Let $s'$ be the value of the arc length for which tangent to $f(s)$ is parallel to the $x_1$ axis (look at the picture above), then
$$\theta_f (s) = \int_{s'}^s k_f(\sigma) d \sigma$$
and, because of the convexity of the arc,
$$- \pi \leq \theta_f(s) \leq \pi$$
The angle
$$\theta_g^* (s) = \theta_g(s) - \theta_g(s') = \int_{s'}^s k_g(\sigma) d \sigma \leq \theta_f(s)$$
Hence
$$d_f = \int_0^L \cos \theta_f (s) ds \leq \int_0^L \cos \theta_g^* (s) ds = d_g \cos \theta_g(s') \leq d_g$$
and $d_f = d_g$ only if $k_f(s) = k_g(s)$ for all $s$. $\square$
My doubts are
Why is true that $\int_0^L \cos \theta_g^* (s) ds = d_g \cos \theta_g(s')$?
How the picture helps me to conclude the assertion of my first doubt?
What is the geometric idea of the proof?
Thanks in advance!
The picture. The picture is wrong. The angles in the picture don't agree with the geometric meaning of $\theta_f$ and $\theta^*_g(s)$. Also, if you look at the curve $f$, the angle $\theta_f(s)$ can lie in $[-3\pi/2, \pi/2]$, and not $[-\pi, \pi]$, as is mentioned in the proof.
Here is a correct picture
The geometric idea of the proof. The first observation is that the angles are defined in terms of the curvature. Secondly, we also know that $d_f = x_{1f}(L) - x_{1f}(0) = \int_0^L \cos \theta_f(s)\,ds$ and $d_g = x_{1g}(L) - x_{1g}(0) = \int_0^L \cos \theta_g(s)\,ds$. Intuitively, we are "adding up" all the $x_1$-components of the tangents along $f$ and $g$. For these reasons, it is naturally to compare "the angle functions on $f$ and $g$".
But which angles should we compare with each other? $\theta_f$ and $\theta_g$? No. Note that $\theta_f$ and $\theta^*_g$ have the same geometric meaning for their respective curves $f$ and $g$. Both $\theta_f$ and $\theta^*_g$ measure the angle from the tangent at $s'$ to the tangent at $s$. That is why we obtain $$ k_g(s) \leq k_f(s) \Rightarrow \theta^*_g(s) \leq \theta_f(s). $$
I think the picture wants to illustrate that we should compare $\theta_f$ and $\theta^*_g(s)$. But it would have been better if the point $g(s)$ was drawn on the arc between $g(s')$ and $g(L)$.
We have choosen $s'$ such that $T_f(s')$ lies in the $x_1$-direction. This makes it easy to get $d_f = \int_0^L \cos\theta_f(s)\,ds$. However, $T_g(s')$ is not necessarily horizontal, and thus we get $d_g\cos \theta_g(s') = \int_0^L \cos\theta^ *_g(s)\,ds$. (See the next paragraph.)
The equality. By the sum formula for cosine we get $$ \begin{align*} \int_0^L \cos \theta^*_{g}(s)\,ds &= \cos\theta_g(s')\int_0^L \cos\theta_g(s)\,ds + \sin \theta_g(s')\int_0^L \sin\theta_g(s)\,ds \\ &= \cos \theta_g(s') \left(x_{1g}(L)-x_{1g}(0)\right) + \sin \theta_g(s') \left(x_{2g}(L)-x_{2g}(0)\right) \\ &= \cos \theta_g(s')\, d_g. \end{align*} $$ You can also get this equality in a geometric way. The integral $\int_0^L \cos\theta^*_g(s)\,ds$ is, by definition of $\theta^*_g(s')$, the length of the $T_g(s')$-component of the segment $x_{1g}(L)-x_{1g}(0)$. (The reasoning is similar to the equation $\int_0^L \cos\theta_g(s)\,ds = x_{1g}(L)-x_{1g}(0)$.) The angle marked with 'v' is $\theta_g(s')$, so this segment has length $d_g \cos\theta_g(s')$.