Circle Theorems: Proofs

 

What Are Circle Theorem Proofs?

• Circle theorem proofs help us prove that certain angle facts in a circle are always true, using logic, geometry rules, and properties of isosceles triangles.
• Many of these proofs start by drawing radii to form isosceles triangles

 

Key Techniques Used in Proofs

• Drawing radii to connect the centre to the circumference
• Using isosceles triangle properties (base angles are equal)
• Using rules like angles in a triangle add to $180\degree$ or angles on a straight line add to $180\degree$

 

1. Proving: The Angle in a Semicircle is $90\degree$

Theorem:

The angle subtended at the circumference of a semicircle is always a right angle.

Proof Steps:

Draw triangle $ABC$ in a circle where $AC$ is a diameter and $B$ is a point on the circumference.

Draw lines from the centre $O$ to points $A$, $B$, and $C$. This forms two isosceles triangles: $OAB$ and $OBC$.

Label the angle at $B$ as $x + y$.

Use isosceles triangle rules to find the angles at $OAB$ and $OBA$ as both $x$, and angles at $OBC$ and $OCB$ as both $y$.

The angles at the centre along the straight line add up to $180$:

$$(180 - 2x) + (180 - 2y) = 180$$

Simplify:

$$360 - 2x - 2y = 180 \Rightarrow x + y = 90\degree$$

So the angle at the circumference is $90\degree$.

 

2. Proving: The Angle at the Centre is Twice the Angle at the Circumference

Theorem:

The angle at the centre of a circle is twice the angle at the circumference subtended by the same arc.

Proof Steps:

Same setup as above.

Let the angle at the circumference be $x + y$, and label the third angle at the centre as $\theta$.

Use the angle around a point rule:

$$\theta + (180 - 2x) + (180 - 2y) = 360$$

Simplify:

$$\theta = 2(x + y)$$

 

3. Proving: Angles in the Same Segment Are Equal

Theorem:

Angles on the circumference from the same arc are equal.

Proof Steps:

Draw radii from the centre to the ends of the arc.

Let the angle at the centre be $2x$.

Then by the previous theorem:

$$\angle ABC = \angle ADC = x$$

So angles in the same segment are equal.

 

4. Proving: Opposite Angles in a Cyclic Quadrilateral Add to $180$

Theorem:

Opposite angles in a cyclic quadrilateral add up to $180$.

Proof Steps:

Draw radii from the centre to opposite vertices.

Let those angles at the centre be $2x$ and $2y$.

Since they lie around a point:

$$2x + 2y = 360 \Rightarrow x + y = 180\degree$$

And those are the angles at the circumference.

 

5. Proving: The Perpendicular from the Centre Bisects a Chord

Theorem:

A perpendicular from the centre of a circle to a chord bisects the chord.

Proof Steps:

1. Draw a radius from the centre that meets the chord at $90\degree$.
2. This creates two right-angled triangles.
3. Use RHS (Right angle, Hypotenuse, Side) to prove the triangles are congruent.
4. Since the triangles are congruent, their corresponding sides (half-chords) are equal.

 

Worked Example

Question:

In the circle below, points $A, B, C$ lie on the circumference and line $DCE$ is a tangent.

Prove that:

$$\angle BCE = \angle BAC$$

Step 1: Draw Diameter

• Draw line $OC$ and extend it across the circle to point $F$, forming diameter $OF$.
• Join points $F$ and $B$ to form triangle $CBF$.

 

Step 2: Use Angle in a Semicircle

$CFB$ is a triangle in a semicircle:

$$\angle CBF = 90\degree$$

 

Step 3: Use Tangent & Radius

Radius $OC$ meets tangent $DE$ at $90\degree$.

Let:

$$\angle BCE = x \Rightarrow \angle FCB = 90 - x$$

 

Step 4: Use Triangle Angle Sum

$$\angle CFB = 180 - 90 - (90 - x) = x$$

 

Step 5: Use Angles in the Same Segment

$\angle BAC$ and $\angle CFB$ are in the same segment:

$$\angle BAC = x = \angle BCE$$

Final Answer:

$$\boxed{\angle BAC = \angle BCE}$$