Power of a point proof I found the question on page 13 of this link.

Let $P$ be a point inside a circle such that there exist three chords through $P$ of equal length. Prove that $P$ is the center of the circle.

I have read one proof related to the power of a point. It went like this: Let $AB$, $CD$ and $EF$ be the three chords also take $AP=a$, $BP=b$, $CP=c$, $DP=d$, $EP=e$, $FP=f$. Using the power of a point we get $ab=cd=ef$. Given that $a+b=c+d=e+f$ (this is where I don't understand), clearly, $\{a,b\}=\{c,d\}=\{e,f\}$. Without loss of generality assume $a=c=e$. Then with $P$ as the centre we draw a circle of radius $a$; then the points $A$, $C$, $E$ coincide with the first circle. Which proves that $P$ has to be the centre.
I apologise for some lines after the "(This is where I don't...)" where I may have skipped some details or if there is an error please do enlighten me. So, what does $\{a,b\}=\{c,d\}=\{e,f\}$ mean? It's not obvious to me that $a=c=e$ also stating "without loss of generality". Thank you.
 A: We know that $ab = cd$ by the power of the point $P$, and also that $a + b = c + d$ because the lengths of the chords $AB$ and $CD$ are equal. So
$$
\begin{align}
&b = \frac{cd}{a} \\
\implies &a + \frac{cd}{a} = c + d \\
\implies &a^2 + cd = ac + ad \\
\implies &a^2 - ac + cd - ad = 0 \\
\implies &(a-c)(a-d) = 0 \\
\implies &a = c\quad \text{OR}\quad a = d
\end{align}
$$
We obtain that $a$ must be equal to one of $c$ and $d$ (possibly both), and hence $b$ must be equal to the other. This is written somewhat confusingly as $\{a,b\} = \{c,d\}$. By similar reasoning we can show the same thing holds for $e$ and $f$ giving $\{a,b\} = \{c,d\} = \{e,f\}$.
(Intuitively, we have two equations relating two unknowns -- the power of point $P$ and the length of the chords -- so there is usually a definite relation between the two. In this case, it is that of equality.)
By naming the points properly, we can always name them such that $a = c = e$ in the above, so we assume this without loss of generality. The rest of the proof follows.
A: Perhaps this can make things easier.

Let $AP = a, BP = b, CP = c$ and $AA’ = BB’ = CC’ = L$. Then, $PA’ = L – a, PB’ = L – b,  PC’ = L – c$.
Note that we can jump to the conclusion directly if $a = b = c$.
The easiest case is when $a, b, c$ are distinct.
We choose to prove the case when $a = c$ but $c \ne b$ and therefore $a \ne b$.
For the chords $AA’$ and $BB’$, we have $a(L-a) = b(L- b)$
After simplification and cancellation (allowable because $a \ne b$), we have $L = a + b$. This gives $PB = b = L-a = PA’$.
Similarly, by considering chords $BB’$ and $CC’$, we have $PB = b = L-c = PC’$.
From the above, we have found a point $P$ (which is inside a circle) such that $PB = PA’ = PC’$ where $B, A’, C’$ are points on the circumference of that circle. This means the only location of $P$ is the center.
