Thanks are due to Sanjay Joshi of The Perse School, Cambridge for this solution.

We shall prove that the line $AOB$ always divides the total perimeter into two equal parts, both of length $\pi r$, defining $r$ to be the radius of the large semicircle as in the diagram. |

First consider the case where the line $AOB$ is horizontal.
The perimeter of the bottom half (the large semicircle) is $\pi r$.
Twice the perimeter of each of the small upper semicircles is $2
{\pi r \over 2}$, again $\pi r$. Hence when the line $AOB$ is
horizontal, the section of the perimeter above the line and the
section below the line are of equal length. As the line $AOB$
rotates about $O$, the lengths of the two sections of the perimeter
change to $\pi r + P - Q$ on one side of the line and $\pi r + Q -
P$ on the other (where $P$ and $Q$ are lengths as defined in the
diagram above). $$Q = r \theta $$ where $\theta$ in this equation
is the angle shown in the diagram, measured in radians. Using the
property that the angle at the centre of a circle is twice the
angle at the circumference subtended by the same arc, $$P = 2\theta
({r \over 2})= r\theta. $$ It is therefore clear that $$P=Q$$ and
that the perimeter length will always be $\pi r$.