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8.04 Chords and arcs

Lesson

When solving geometric problems inside circles, there are some useful relationships between angles and distances relating to equal chords or arcs.

 

Properties of equal chords

When we have chords of equal length, we can demonstrate relationships between them by constructing angles that are subtended by them.

Doing this allows us to find congruent triangles and the corresponding information between them.

Summary

When a chord subtends an angle, it means that the chord and the two arms of the angle form a triangle where the chord lies opposite to the angle.

When an arc subtends an angle, we instead get a sector with the arc opposite the sector's angle.

Exploration

The diagram below shows two equal chords on a circle, joined to the centre by some radii.

Since radii in a circle are equal, we know that $OA$OA, $OB$OB, $OC$OC and $OD$OD are all equal. We can write this information as:

$OA=OC$OA=OC (Radii of a circle are equal)

$OB=OD$OB=OD (Radii of a circle are equal)

Since we are given that the chords are equal, we also know that:

$AB=CD$AB=CD (Given)

This is enough information to determine that $\triangle OAB$OAB and $\triangle OCD$OCD are congruent under the test SSS, since we have shown that they have three pairs of equal sides.

As a result of the two triangles being congruent, we know that $\angle AOB=\angle COD$AOB=COD since they are corresponding angles in congruent triangles.

Since this will work for any two equal chords, we have proved that:

Summary

Chords of equal length subtend equal angles at the centre.

Continuing with our exploration, we can draw perpendicular lines from the centre to each chord, like so:

In our four triangles, we know that:

  • $AO$AO, $BO$BO, $CO$CO and $DO$DO are all equal since they are radii
  • $\angle OAX$OAX, $\angle OBX$OBX, $\angle OCY$OCY and $\angle ODY$ODY are all equal since they are base angles in congruent isosceles triangles
  • $\angle OXA$OXA, $\angle OXB$OXB, $\angle OYC$OYC an $\angle OYD$OYD are all right angles

This is enough information to show that all four triangles are congruent under the test AAS.

As a result of these triangles being congruent, we now know that:

  • $OX$OX and $OY$OY must be equal since they are corresponding sides
  • $AX$AX, $BX$BX, $CY$CY and $DY$DY are all equal since they are also corresponding sides

Thus, we have now proved that:

Summary

Chords of equal length are equidistant from the centre.

Summary

The perpendicular line from a centre of a circle to a chord bisects its chord.

Since any perpendicular line through the centre will be a bisector of the chord, we find that any two of the three factors will guarantee the third. What does this mean?

It means that we have also proved that:

Summary

The line from the centre of a circle to the midpoint of a chord is perpendicular to the chord.

Summary

The perpendicular bisector of a chord must pass through the centre of the circle.

Another relationship between chords and angles can be found when we consider the angle at the centre compared to the angle at the circumference.

Exploration

In the diagram below, we have an angle at the centre and an angle at the circumference subtended by the same chord.

Since they are all radii, we know that $OA$OA, $OB$OB and $OC$OC are all equal and that $\triangle OAC$OAC and $\triangle OBC$OBC are both isosceles.

Given that $\angle OCA=x$OCA=x, then $\angle OAC=x$OAC=x (base angles in isosceles triangles are equal).

Similarly, given that $\angle OCB=y$OCB=y, then $\angle OBC=y$OBC=y.

Since exterior angles are equal to the sum of the two opposite interior angles, we can find that:

$\angle AOP=2x$AOP=2x

$\angle BOP=2y$BOP=2y

Collecting our angles, we have shown that:

$\angle ACB=x+y$ACB=x+y

$\angle AOB=2x+2y$AOB=2x+2y

In other words:

$\angle AOB=2\angle ACB$AOB=2ACB

Since we move the point $C$C around the circle freely, we have proved that:

Summary

Angle at the centre of a circle is twice the angle at its circumference.

It is worth noting that if we move the point $C$C too far around the circle, the point $P$P will no longer lie on the arc $AB$AB. In order to prove our angle relation for this case, we need to construct an additional line, like so:

If we let $\angle OCB=x$OCB=x and $\angle ACB=y$ACB=y, we can use a similar proof except that it will involve subtraction instead of addition.

Explore this proof using the applet below:

 

Angles on equal chords and arcs

Since every chord has a matching arc, the relation between angles and arcs is the same as the ones we proved between angles and chords.

This means that an angle at the centre will still be twice the angle at the circumference, even if its subtended by an arc rather than a chord.

We can extend this idea to show that any two angles standing on the same chord or arc must be equal.

Exploration

Consider these angles subtended by the same arc.

Consider that angle $\angle AOB$AOB is the angle at the centre with respect to $\angle ADB$ADB, we can see that $\angle AOB=2\angle ADB$AOB=2ADB.

Also, $\angle AOB$AOB is the angle at the centre with respect to $\angle ACB$ACB, so $\angle AOB=2\angle ACB$AOB=2ACB.

Since $\angle ADB$ADB and $\angle ACB$ACB are both equal to half $\angle AOB$AOB, they must be equal.

Summary

Angles standing on equal length chords or arcs are equal.

Equivalently, angles in the same segment are equal.

 

Angles in the semicircle

Now that we know that an angle at the centre is twice an angle at the circumference (when standing on the same chord), we can consider the particular case where the chord is the diameter.

In this case, we can see that the angle at the centre is a straight angle. In other words, the angle at the centre is $180^\circ$180°.

Since the angle at the circumference must be equal to half of this, we find that any angle standing on the diameter must be a right angle.

Summary

The angle in a semicircle is a right angle.

 

Practice questions

Question 1

In the diagram, $O$O is the centre of the circle. Solve for $x$x.

  1. Show all working and reasoning.

Question 2

In the diagram, $\angle BAC=53^\circ$BAC=53°. Solve for $x$x.

  1. Show all working and reasoning.

Question 3

In the following diagram, $O$O is the centre of a circle with $AB=CD$AB=CD.

  1. Prove that $\triangle ABO$ABO and $\triangle CDO$CDO are congruent.

    Show all working and reasoning.

  2. Hence, or otherwise, prove $\triangle BEO$BEO and $\triangle DFO$DFO are congruent

    Show all working and reasoning.

Outcomes

VCMMG366 (10a)

Prove and apply angle and chord properties of circles.

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