Consider circle $\bigcirc O$ with chord $AB$ which is not a diameter. Let $P$ be a point on $AB$ which is not the midpoint. Clearly there exists a chord $CD$ which is congruent to $AB$ and intersects it at $P$. (Proof: Reflect $A$ over diameter $PO$ to find $C$ and $B$ over $PO$ to find $D$.)
I believe this chord $CD$ is unique, but can't find a simple proof of it. Below is my (complicated) proof. What is a simple way to prove uniqueness of $CD$?
Assume WLOG $$AP < PB \\ CP \leq PD.$$ By Power of Point $P$, we have $$AP \cdot PB = CP \cdot PD.$$ If $$AP + PB = CP + PD$$ then $AP \cong CP$ as shown here.
By Euclid's III.7, there is only one point $C$ on the circle such that $C \neq A$ and $AP \cong CP$. This completes the proof, but seems needlessly complicated*. Is there a simpler proof?
*Why? The claim seems to be a simple observation necessitated by symmetry (or perhaps also noting that segment length is a monotonic function of chord angle, III.7). Instead, I had to use the full power of the PoP theorem, combined with quadratic algebra (needed to show $ab=cd, a+b=c=d \implies \{ a,b \} = \{ c,d \}$). That's simply too much machinery for a simple observation.