Circle chord theorems

Here you will learn about circle theorems involving the chords of a circle and the relationships they create. You will learn the application of these theorems, the proofs of these theorems, and how to use them to solve more complex problems.

Students first learn how to work with these circle theorems in a geometry course where they spend a time delving into the proofs and application. Circle theorems are essential for success on graduation assessments as well as college entrance assessments.

What are circle chord theorems?

Chord circle theorems are geometrical properties of circles that can be calculated using the chord of a circle. The chord of a circle is a line segment that connects two points that sit on the edge of a circle. The longest chord in a circle is the diameter of the circle.

Recall the parts of a circle that are essential to understand circle chord theorems.

• The diameter of the circle is the width of a circle, through the center. The endpoints of the diameter lie on the circumference of the circle. The diameter is twice the length of the radius of the circle.

• An arc is a part of the circumference.

• The major segment is the larger part of a circle when it is enclosed by a chord and the major arc.

• The minor segment is the smaller part of a circle when it is cut by a chord and the minor arc.

• The radius of a circle is the distance between the center of the circle and the circumference of the circle.

Let’s take a look at the first circle theorem.

1) Diameter perpendicular to a chord

The perpendicular from the center of a circle to a chord bisects the chord (splits the chord into two equal parts).

In the diagram above, AB is a chord and CE is a radius (half the diameter).

Here we can see that the radius CE bisects the chord AB at 90 degrees and the lengths AD=BD=x

You can use congruent triangles to prove this is always true.

First, draw two radii, AC and BC, to make two right triangles, triangle ADC and triangle BDC.

Since radii within a circle are congruent in length, you know that the hypotenuses of both triangles are congruent.

Since both triangles share side CD, through the reflexive property, CD is congruent to itself.

Therefore, using the triangle congruence theorem of Hypotenuse-Leg, you can conclude that triangle ADC is congruent to triangle BDC.

Since the two triangles are congruent, you can further conclude through the definition of congruent triangles that AD is congruent to BD.

This proves that when the diameter (or radius) is perpendicular to a chord, it also bisects (cuts in half) the chord.

2) Congruent chords

Chords in a circle are congruent if and only if they are the same distance from the center of the circle.

In circle B, BO and BP are the same length, x units.

Therefore, chord LM and chord NQ are congruent because they are the same perpendicular distance from the center of the circle.

Again, you can use congruent triangles to prove this is true.

First, sketch in the radii to create two triangles. All the radii within the same circle are congruent. Triangle LBM and triangle NBQ are isosceles triangles.

Using the Isosceles triangle theorem, you know that opposite congruent sides of a triangle are congruent angles.

Angle BNQ is congruent to angle BQN. Also, angle BML is congruent to angle BLM.

Since BN is congruent to BQ which is congruent to BM which is congruent to BL, you can conclude that angle BNQ is congruent to angle BQN which is congruent to angle BML which is congruent to angle BLM.

Hence, using the Angle-Angle-Side triangle congruent theorem, you can conclude that triangle NBQ is congruent to triangle LBM.

Since the triangles are congruent, then using the definition of congruent triangles, you can conclude that side NQ is congruent to side LM.

3) Congruent chords and intercepted arcs

When two chords in a circle are congruent, the arcs they intercept are congruent.

If chord JK is congruent to chord GH, then arc JK is congruent to arc GH.

You can use central angles to prove this is true.

First, draw in two diameters.

The diameters create central angles. Central angle GFH is congruent to angle JFK because they are vertical angles.

Angle GFH intercepts arc GH and angle JFK intercepts arc JK.

Since angle GFH is congruent to angle JFK, you can conclude that arc GH is congruent to arc JK because central angles are equal in measure to the arcs they intercept.

Likewise, angle GFJ is congruent to angle HFK because they are vertical angles. So, arc GJ must be congruent to arc HK.

Note that for this example, the two arcs GH and JK are parallel. If they are not parallel, the proof would be similar to that for “(2) Congruent Chords ”.

Subtended angles

An angle within a circle is created by two chords meeting at a point on the circumference. The diagrams below show the inscribed angle subtended by arc AC from point B for two different circles.

Top tip: The word subtend is used a lot within circle theorems so make sure you know what it means.

What are circle chord theorems?

Common Core State Standards

How does this apply to geometry?

• High School Geometry – Circles (HSG-C.A.2)
  Identify and describe relationships among inscribed angles, radii, and chords. Include the relationship between central, inscribed, and circumscribed angles; inscribed angles on a diameter are right angles; the radius of a circle is perpendicular to the tangent where the radius intersects the circle.

How to use circle chord theorems to solve problems

In order to find missing angles or the length of a chord:

1. Identify the key parts of the circle that are given.
2. Apply the appropriate theorem to problem solve.
3. Describe or create an equation to find the missing part.
4. Find the solution.

Circle chord theorems examples

Example 1:

In circle C, DK is perpendicular to GH. If GH is 26 units, find HJ.

1. Identify the key parts of the circle that are given.

Since C is the center of the circle, DK is the diameter, which means that CK is the radius.

GH is a chord that has a length of 26 units.

2Apply the appropriate theorem to problem solve.

Using the theorem that if the diameter or radius is perpendicular to a chord, then it bisects the chord – meaning it cuts the chord into two equal halves. So, the diameter is the perpendicular bisector of a chord.

3Describe or create an equation to find the missing part.

In this case, divide 26 by 2 or multiply it to \cfrac{1}{2}.

4Find the solution.

HJ=13 units

Example 2: triangle

In circle A, diameter BC is perpendicular to DF. DF is equal to 8 and BC=10, find the length of BE.

Identify the key parts of the circle that are given.

Diameter BC is perpendicular to chord DF so DF is also being bisected by BC. AB and AC are radii of the circle. As the radius of any circle is half the length of the diameter, AB=AC=5.

Apply the appropriate theorem to problem solve.

Using the theorem that if the diameter or radius is perpendicular to a chord, then it bisects the chord – meaning it cuts the chord into two equal halves.

DE=EF=8\div{2}=4

Describe or create an equation to find the missing part.

In this case, draw in another radius to create a right triangle AEF.

The radius is the hypotenuse of the triangle, AF=5.

EF is one of the legs of the right triangle, EF=4.

Since it is a right triangle, use the Pythagorean Theorem (a^{2}+b^{2}=c^{2}), find the missing leg, AE.

Find the solution.

Let AE=a, EF=b=4, and AF=c=5 (remember c is always the hypotenuse of a right triangle).

Then,

\begin{aligned} a^{2}+4^{2}&=5^{2} \\\\ a^{2}+16&=25 \\\\ a^{2}&=25-16=9 \\\\ a&=\sqrt{9}=3 \end{aligned}

So, AE=3 units.

Since the radius AB is 5 units, subtract AE from AB to find BE,

AB-AE=5-3=2.

BE=2 units.

Example 3: arcs

In circle B, chord LM is congruent to chord KN.

Minor arc LM=5x-2

Minor arc KN=3x+14

Find the measure of Minor arc KN.

Identify the key parts of the circle that are given.

In circle B, chord LM intercepts minor arc LM and chord KN intercept minor arc KN.

Find the solution.

Substitute x=8 into 3x+14 to find the length of arc KN\text{:}

Arc KN=3(8)+14=38 units

Example 4:

In circle M, LK is the diameter and perpendicular to PR. If RS=7x-3 and PS=3x+17, find PR.

Identify the key parts of the circle that are given.

In circle M, KL is a diameter and is perpendicular to PR. \; KM is a radius of circle M.

RS=7x-3

PS=3x+17

Find the solution.

Substitute x=5 into 7x-3 to find the measure of RS:

RS=7(5)-3=32

Substitute x=5 into 3x+17 to find the measure of PS\text{:}

PS=3(5)+17=32

Find the sum of RS and PS to find the measure of PR:

PR=RS+PS=32+32=64 units

Example 5:

In circle L,

• QU=5a

• Chord ST=60

• LU=LW=4 units

• Angle LWT= Angle LUQ=90^{\circ}.

Find the value of a.

Identify the key parts of the circle that are given.

QU=5a

Chord ST=60

LU=LW=4

Example 6: semicircle

In circle B, minor arc LM and minor arc KN are congruent to each other. Chord LM=x^{2} and chord KN=x+6, find the measure of LM.

Find the solution.

As this is a quadratic, rearrange the quadratic to the general form ax^2+bx+c=0\text{:}

This quadratic can be solved by factorising as b^2-4ac=25

Step-by-step guide: Factoring quadratics

\begin{aligned} &x^{2}-x-6=0\\\\ &(x-3)(x+2)=0\\\\ &x=3\text{ and }x=-2 \end{aligned}

Check the values to see if they work in the original problem. Remember, lengths of chords cannot be negative.

When x=3, LM=x^{2}=3^{2}=9>0.

When x=3, KN=x+6=3+6=9>0.

(LM=KN>0)

When x=-2, LM=x^{2}=(-2)^{2}=4>0

When x=-2, KN=x+6=-2+6=4>0.

(LM=KN>0)

Both values of x work in this case.

Teaching tips for circle chord theorems

• Provide opportunity for students to discover the theorems and make conjectures using compasses and protractors.
  
• Instead of worksheets, have students use online platforms for game playing to practice skills.
  
• Incorporate digital platforms such as Desmos for students to explore and investigate the theorems.

Easy mistakes to make

• Forgetting to sketch in a radius to create a triangle
  In certain cases, it is helpful to draw in a radius to create a right triangle to find a missing chord length. For example, to find the length EF, it might be helpful to make a right triangle.
  
  

  

• Not knowing which theorem to apply
  Building conceptual understanding of the theorems through constructions helps students to remember and apply the theorems. When problem solving, sometimes you have to apply more than one theorem to solve.

Related circle theorems Lessons

• Central angles
• Chords of a circle
• Subtended
• Tangent of a circle

The next lessons are

• Types of data
• Averages and range
• Representing data

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