# Moving chord on the simple closed curve

Consider a simple closed curve $C$ in $\mathbb{R}^2$. For any points $a$ and $b$ on this curve we associate point $c$ on the left (or right) side to chord $ab$ such that $\angle acb = 90^{\circ}, ac=cb.$ Continuously moving (sliding) a chord $ab$ in such a way that it goes to $ba$ (and $a(t)$ never equals $b(t)$, i.e. $\|a(t)-b(t)\|>0 ~\forall t$), the point $c$ will draw a line $L$. My hypothesis is that for any such trajectory and for any $C$, $L$ will intersect $C$. Is it true?

My intuition says that curve $L$ can never be completely inside or outside $C$, but I don't now how to prove it, because geometry is not not my domain.

P.S. Curve $C$ can be non smooth.

• Surely for the outside case you can just take a circle and keep the length of the chord $ab$ small and fixed? – James Smith Aug 14 '17 at 20:39