Chord Properties

Draw a circle with centre $O$ and any chord $AB$ that is not a diameter. Now draw $OA$ and $OB$. What kind of triangle is $\triangle OAB$, and why must it always be that kind, no matter which chord you chose? Write the reason down before reading on.

Every chord property comes from one move: join the centre to both ends of the chord. Because all radii of a circle are equal, that always produces an isosceles triangle, and every chord theorem is a property of isosceles triangles restated in the language of circles.

Given: a circle with centre $O$ and chord $AB$, with $OM \perp AB$.
To prove: $AM = MB$.

Note the test. $OA$ and $OB$ are the hypotenuses, opposite the right angles, and $OM$ is a matching pair of other sides. SSS is not available, because $AM = MB$ is exactly what is being proved and cannot be assumed.

The converse says: if a line from the centre bisects a chord, it is perpendicular to it. It needs its own proof, and it gets one from SSS rather than RHS.

With $AM = MB$ given, $OA = OB$ as radii and $OM$ common, the triangles are congruent by SSS. So $\angle OMA = \angle OMB$, and since they are adjacent angles on the straight line $AB$ they sum to $180°$. Each is therefore $90°$.

Putting the two directions together gives a third statement worth having: the perpendicular bisector of any chord passes through the centre.

That result is more useful than it looks. It means the centre of a circle can be found from the circle alone: draw any two chords, construct their perpendicular bisectors, and the point where they meet is the centre. This is how you locate the centre of a circular object with no marked middle.

Given: chords $AB$ and $CD$ with $AB = CD$, and $OM \perp AB$, $ON \perp CD$.
To prove: $OM = ON$.

The previous theorem does the work. Since $OM \perp AB$, it bisects it, so $AM = \tfrac{1}{2}AB$. Similarly $CN = \tfrac{1}{2}CD$. Because $AB = CD$, the halves are equal: $AM = CN$.

Now in $\triangle OMA$ and $\triangle ONC$: $OA = OC$ (radii), $AM = CN$ (just shown), and both triangles have a right angle at $M$ and $N$. By RHS they are congruent, so $OM = ON$.

The converse is also true and proved the same way in reverse: chords equidistant from the centre are equal. Together they say that in one circle, the length of a chord and its distance from the centre determine each other, so a longer chord is always closer to the centre.

Once the chord is bisected, the diagram contains a right-angled triangle with three named lengths: the radius as hypotenuse, half the chord as one leg, and the distance from the centre as the other.

$$r^2 = d^2 + \left(\frac{\text{chord}}{2}\right)^2$$

Any two of the three give the third. If a circle of radius $13$ cm has a chord $24$ cm long, the half-chord is $12$, so $13^2 = d^2 + 12^2$, giving $d^2 = 169 - 144 = 25$ and $d = 5$ cm.

Two habits make this reliable. Halve the chord first, because forgetting to is the commonest error and gives an impossible answer. And check the size: the distance must be less than the radius, and the half-chord must also be less than the radius, so an answer larger than $r$ signals an error immediately.

Four quick problems. Work each one, then reveal the answer.

1. 1 A radius of $10$ cm, a chord $2$ cm from the centre. Find the half-chord.

   Show answer
   $10^2 = 2^2 + x^2$, so $x^2 = 96$.$x = \sqrt{96} \approx 9.8$ cm

2. 2 In one circle, chord $P$ is longer than chord $Q$. Which is closer to the centre?

   Show answer
   Length and distance determine each other; longer means closer.Chord $P$

3. 3 A line from the centre bisects a chord. What angle does it make with the chord?

   Show answer
   This is the converse theorem, proved by SSS.$90°$

4. 4 Radius $25$ cm, chord $48$ cm. Find the distance from the centre.

   Show answer
   Half-chord $24$, so $d^2 = 625 - 576$.$7$ cm