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10&10a

8.03 Chords and arcs

Lesson

Introduction

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.

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.

A circle with center O and triangles A O B and C O D inside the circle. Line segments D C and A B are chords.

The diagram 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, \, OB, \, OC and OD are all equal. We can write this information as:\text{ } \\ \begin{array}{cll} OA = OC &\text{(Radii of a circle are equal)} \\ OB = OD &\text{(Radii of a circle are equal)} \\ \end{array}

Since we are given that the chords are equal, we also know that:\text{ } \\ \begin{array}{cll} AB = CD &\text{(Given)} \\ \end{array}

So \triangle OAB \equiv \triangle OCD by SSS.

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

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

A circle with center O and triangles A O B and C O D inside the circle. Ask your teacher for more information.

In our four triangles, we know that:

  • AO, \, BO, CO and DO are all equal radii.

  • \angle OAX, \, \angle OBX, \, \angle OCY and \angle ODY are all equal since they are base angles in congruent isosceles triangles.

  • \angle OXA, \, \angle OXB, \, \angle OYC and \angle OYD are all right angles.

So all four triangles are congruent under the test AAS.

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

  • OX and OY must be equal since they are corresponding sides.

  • AX, \, BX, \, CY and DY are all equal since they are also corresponding sides.

We have now proved that:

  • Chords of equal length are equidistant from the centre.

  • 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:

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

  • 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.

A circle with center O and triangles A O C and C O B inside. A C and C B are chords. Ask your teacher for more information.

In the diagram, 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, \, OB and OC are all equal and that \triangle OAC and \triangle OBC are both isosceles.

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

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

Since exterior angles are equal to the sum of the two opposite interior angles, we can find that:\angle AOP = 2x \\ \angle BOP = 2y

Collecting our angles, we have shown that:\angle ACB = x + y \\ \angle AOB = 2x + 2y

In other words:\angle AOB = 2 \angle ACB

Since we move the point C around the circle freely, we have proved that: The angle at the centre of a circle is twice the angle at its circumference.

A circle with chords C A and C B, radii O A and O B and diameter C P is dashed. Angle O C B is x, and angle ACB is y.

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

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

Exploration

Explore the above proof using the applet. Move the points on the circle.

Loading interactive...

The angle at the centre of a circle is always twice the angle at its circumference that stands on the same arc.

Examples

Example 1

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

Circle with 2 angles standing on the same arc. The angle at the centre is 74 and the angle at the circumference is x degrees
Worked Solution
Create a strategy

Use the fact that the angle at the centre is double the angle at the circumference.

Apply the idea
\displaystyle 2x\displaystyle =\displaystyle 74(Angle at the centre of a circle is twice the angle at its circumference)
\displaystyle x\displaystyle =\displaystyle 37Divide both sides by 2

Example 2

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

Circle with centre O. OE is perpendicular to AB of triangle A B O. OF is perpendicular to CD of triangle C D O.
a

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

Worked Solution
Create a strategy

Show that the triangles satisfy one of the congruence tests: SSS, AAS, RHS, or SAS.

Apply the idea
To prove: \triangle ABO \equiv \triangle CDO
StatementsReasons
1.AB=CD(Given)
2.AO=CO, \, BO=DO(Equal radii)
3.\triangle ABO \equiv \triangle CDO(SSS)
b

Prove \triangle BEO and \triangle DFO are congruent.

Worked Solution
Create a strategy

Show that the triangles satisfy one of the congruence tests: SSS, AAS, RHS, or SAS.

Apply the idea
To prove: \triangle BEO \equiv \triangle DFO
StatementsReasons
1.\angle EBO = \angle FDO(Corresponding angles in congruent triangles)
2.\angle BEO = \angle DFO(Given)
3.BO=DO(Equal radii)
4.\triangle BEO \equiv \triangle DFO(AAS)
Idea summary

Properties of equal chords:

  • Chords of equal length subtend equal angles at the centre.

  • Chords of equal length are equidistant from the centre.

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

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

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

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

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.

Equal angles A C B and A D B stand on the arc A B and are at the circumference. Ask your teacher for more information.

Consider these angles subtended by the same arc.

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

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

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

Angles standing on equal length chords or arcs are equal.

Equivalently, angles in the same segment are equal.

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.

Circle with centre O. Triangle A B C is made up of diameter A B, chord A C and chord C B. Angle A C B is right angled.

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\degree.

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.

The angle in a semicircle is a right angle.

Examples

Example 3

In the diagram, \angle BAC = 45\degree. Solve for x.

A circle with two triangles, A D C and A B C.  Angle A D C is x, angle B A C is 45 degrees, and angle ACD is right.
Worked Solution
Create a strategy

Use the property that angles in the same segment are equal.

Apply the idea
\displaystyle \angle ABC\displaystyle =\displaystyle x(Angles in the same segment)
\displaystyle x+45+90\displaystyle =\displaystyle 180(Angle sum of a triangle)
\displaystyle x+135\displaystyle =\displaystyle 180Add like terms
\displaystyle x\displaystyle =\displaystyle 45Subtract 135 from both sides
Idea summary

Angle properties:

  • Angles standing on equal length chords or arcs are equal.

  • Angles in the same segment are equal.

  • The angle in a semicircle is a right angle.

Outcomes

ACMMG272 (10a)

Prove and apply angle and chord properties of circles

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