:400px Let $P = \left({x, y}\right)$ be a general point on $C$. Let $Q = \left({X, Y}\right)$ be the center of curvature of $C$ at $P$. From the above diagram: :$(1): \quad \begin{cases} x - X = \pm \rho \sin \psi \\ Y - y = \pm \rho \cos \psi \end{cases}$ where: :$\rho$ is the radius of curvature of $C$ at $P$ :$\psi$ is the angle between the tangent to $C$ at $P$ and the $x$-axis. Whether the sign is plus or minus depends on whether the curve is convex or concave. For simplicity, let it be assumed that the curvature $k$ at each point under consideration on $C$ is positive. The case for $k < 0$ can then be treated similarly. Thus we have $k > 0$ and so $(1)$ can be written: :$(2): \quad \begin{cases} X = x - \rho \sin \psi \\ Y = y +\rho \cos \psi \end{cases}$ By definition of curvature: :$k = \dfrac {\mathrm d \psi} {\mathrm d s}$ and: :$\rho = \dfrac 1 k = \dfrac {\mathrm d s} {\mathrm d \psi}$ Hence: {{begin-eqn}} {{eqn | l = \rho \sin \psi       | r = \dfrac {\mathrm d s} {\mathrm d \psi} \dfrac {\mathrm d y} {\mathrm d s}       | c =  }} {{eqn | n = 3       | r = \dfrac {\mathrm d y} {\mathrm d \psi}       | c =  }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \rho \cos \psi       | r = \dfrac {\mathrm d s} {\mathrm d \psi} \dfrac {\mathrm d x} {\mathrm d s}       | c =  }} {{eqn | n = 4       | r = \dfrac {\mathrm d x} {\mathrm d \psi}       | c =  }} {{end-eqn}} Differentiating $(2)$ {{WRT|Differentiation}} $\psi$: {{begin-eqn}} {{eqn | l = \dfrac {\mathrm d X} {\mathrm d \psi}       | r = \dfrac {\mathrm d x} {\mathrm d \psi} - \rho \cos \psi - \dfrac {\mathrm d \rho} {\mathrm d \psi} \sin \psi       | c =  }} {{eqn | n = 5       | r = - \dfrac {\mathrm d \rho} {\mathrm d \psi} \sin \psi       | c = from $(3)$ }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \dfrac {\mathrm d Y} {\mathrm d \psi}       | r = \dfrac {\mathrm d y} {\mathrm d \psi} - \rho \sin \psi + \dfrac {\mathrm d \rho} {\mathrm d \psi} \cos \psi       | c =  }} {{eqn | n = 6       | r = \dfrac {\mathrm d \rho} {\mathrm d \psi} \cos \psi       | c = from $(4)$ }} {{end-eqn}} Let $S$ be the length of arc of the evolute $E$ from a fixed point $Q_0$ on $E$ to the variable point $Q$ corresponding to $P$. We have that: :$\mathrm d S^2 = \mathrm d X^2 + \mathrm d Y^2$ and so: {{begin-eqn}} {{eqn | l = \left({\dfrac {\mathrm d S} {\mathrm d \psi} }\right)^2       | r = \left({\dfrac {\mathrm d X} {\mathrm d \psi} }\right)^2 + \left({\dfrac {\mathrm d Y} {\mathrm d \psi} }\right)^2       | c =  }} {{eqn | r = \left({- \dfrac {\mathrm d \rho} {\mathrm d \psi} \sin \psi}\right)^2 + \left({\dfrac {\mathrm d \rho} {\mathrm d \psi} \cos \psi}\right)^2       | c = from $(5)$ and $(6)$ }} {{eqn | r = \left({\dfrac {\mathrm d \rho} {\mathrm d \psi} }\right)^2 \left({\sin^2 \psi + \cos^2 \psi}\right)       | c =  }} {{eqn | r = \left({\dfrac {\mathrm d \rho} {\mathrm d \psi} }\right)^2        | c = Sum of Squares of Sine and Cosine }} {{end-eqn}} Choosing the direction of $S$ so that $S$ and $\rho$ increase together, this tells us: :$\dfrac {\mathrm d S} {\mathrm d \psi} = \dfrac {\mathrm d \rho} {\mathrm d \psi}$ Integrating {{WRT|Integration}} $\psi$ gives: :$S = \rho + c$ where $c$ is a constant. {{qed}}