Prove the perpendicular bisector of chord passes through the centre of the circle 
Hello, can someone please give me a simple proof to the following theorem:
"The perpendicular bisector a chord passes through the centre of the circle."
I have attached a diagram of what I mean and web link of a proof that I did not understand below.
https://proofwiki.org/wiki/Perpendicular_Bisector_of_Chord_Passes_Through_Center
Please explain simply and fully because I have an exam on this tomorrow. Also, could you explain the converse theorem whereby a bisector passes through the centre of the circle, prove it's perpendicular and a perpendicular line passes through the centre, prove it bisects the chord.

As Doug M said in a comment, the last statement of the “proof” in the question is false: “equal distance from two points on the circumference” is not a unique property of the circle’s centre. In fact every point within a circle lies an equal distance from some two points on the circumference!
Perhaps what they meant to say is that the centre is the unique point at a distance $r$, the radius, from any two points on the circumference. I’ll use this lemma below to complete the proof.

What they’ve actually proven is that the perpendicular bisector is the locus of points equidistant from $A$ and $B$. Point $C$ is an arbitrary point on this locus.
It is necessary now to prove that the circle’s centre is also on this locus, and hence is a possible location for $C$. By the above lemma, this requires that the radius be a possible length for the line segments $CA$ and $CB$.
The minimum length of $CA$ is in the degenerate case where $C$ and $D$ coincide. This length—call it $x$ is anywhere from 0 (if $A$ and $B$ also coincide) to $r$ (if $AB$ is a diameter of the circle). That is, $x\in[0, r]$.
The maximum length of $CA$ is unbounded, so we can say that its length is on the interval $[x, \infty)$. Given the above constraints on $x$, the radius $r$ lies within this interval. QED.

Okay, I did run quickly past one issue: given two points $A$ and $B$ on the circumference, there are two points at a distance $r$ from both of them. One of them is the centre of the circle, but the other is not (unless $AB$ is a diameter), and in fact it may even be exterior to the circle.
I could prove that both of these points lie on the perpendicular bisector… or I could observe that the two points lie one each in the two half-planes bounded by the line $AB$. If we constrain $C$ to lie in the same half-plane as the circle’s centre, then it must be the circle’s centre. QED.
