Prove the perpendicular bisector of chord passes through the centre of the circle

Question

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.

Answer 1

Consider the same figure to which you have linked.

In the figure, draw a line perpendicular to AB at the point B. Say, this intersects the circle at the point E. Let D be the mid point of AB. Join CD. We prove first that $\triangle$ ACD and $\triangle$ AEB are similar.
AC = CE = r (radius of the circle)
AD = DB (since D is the mid point of AB)
So we have,
$$ \frac {AC}{AD} = \frac{AE}{AB} = 2$$ $$\angle CAD = \angle EAB \space (Common)$$ With these conditions, it is clear that these two triangles are similar. So, their corresponding angles are congruent. Specifically, $$ \angle ADC = \angle ABE = 90^0 \space (by \space construction)$$ This proves that the line CD is the perpendicular bisector, C being the center of the circle.

Answer 2

The perpendicular bisector of a segment $[AB]$ is the locus of points $M$ equidistant from $A$ and $B$.

This means $MA=MB$.

But if we set $MA=R$ then this means $A,B$ are on the circle of centre $M$ and radius $R$.

And since the chord in this case is precisely $[AB]$ and $M$ belongs to the perpendicular bisector, you have your result.

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If we define the perpendicular bisector by the line $\perp[AB]$ and passing by $I=\frac{A+B}2$ the middle of $[AB]$.

Then for any point $M$ on the bissector, we can apply pythagoras' theorem.

$\begin{cases} MA^2=MI^2+AI^2\\ MB^2=MI^2+BI^2 \end{cases}$ but since $AI=BI$, because $I$ middle of $[AB]$ then $MA=MB$.

And we can conclude like previously.

Answer 3

Start with a diameter.

Construct the two tangent lines to the circle at the endpoints of the diameter.

Drop the diameter along the tangent lines until it matches with the chord.

By symmetry, the line between the midpoint of the chord to the origin of the circle is perpendicular to the chord.

Answer 4

For first part of question the position of $C$ is at first chosen arbitrarily in the plane. After $SAS$ is satisfied with givens, The triangles $ MAC, MBC $ are congruent, $C$ must lie on the bisector. No matter where the point $C$ is chosen on the perpendicular bisector, the circle through $A,B$ has to have its center somewhere on $MC$ line, it is sufficient for a proof.

Answer 5

The proof in the picture is the simplest possible one. All you have to do is write the conditions for congruence thus proving that the triangles in the above picture are congruent. This is the best possible approach to the given question.

Answer 6

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.

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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.

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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.