# 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.

• What step do you not understand of the proof you linked to? – irh May 24 '17 at 20:00
• the closing statment at the end – Inquirer May 24 '17 at 20:02
• The proof under link "enter image description here" is a false proof, because the concluding statement is not true. – Doug M May 24 '17 at 20:04
• Since the triangles are right triangles, then $DF$ perpendicular to $AB$. Since $D$ by definition is the middle of $AB$, then $DF$ is a perpendicular bisector. – irh May 24 '17 at 20:07
• Prove that the line from the center of the circle to the midpoint of the chord is the perpendicular bisector. And since the perpendicular bisector of any segment is unique the center lies on the perpendicular bisector of the chord. – Doug M May 24 '17 at 20:12

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.

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.

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.

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.

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.

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.

Actually, instead of using SAS criterion, you can use SSS criterion. $$AC = BC$$ [∵ AC and BC are radii] $$CD = BD$$ [∵ BC is bisected to form CD and BD] $$DC = DC$$ [Common]

If the perpendicular bisector does not pass through C, then the hypotenuse sides would not be radii, therefore the triangles won't be congruent, therefore the bisector won't be perpendicular, which is a proof by contradiction.