Central and Inscribed Angles

The Arrowhead (or Dart): The most common representation, where the lines form a shape pointing towards the arc.

The "angle in a semicircle is 90°" rule is actually just a specific case of this theorem. If the central angle is a straight line ($180^\circ$), the corresponding inscribed angle at the circumference must be half of that ($90^\circ$).

Proving the Theorem

OCR mark schemes frequently require you to write a formal algebraic proof for this theorem. You must use specific geometric properties to build your case step-by-step.

Step 1: Set up the geometry

Draw a circle with centre $O$. Place a point $A$ on the top of the circumference, and points $B$ and $C$ at the bottom to form an arc. Draw straight lines joining $A$, $B$, and $C$ to the centre $O$. Finally, extend the line $AO$ straight through the centre to a new point $D$ on the opposite side.

Step 2: Identify equal lengths

Every line drawn from the centre $O$ to the circumference ($OA$, $OB$, and $OC$) is a radius. Because all radii in a single circle are equal in length, $OA = OB = OC$.

Step 3: Find the isosceles triangles

Because triangle $OAB$ has two equal sides ($OA$ and $OB$), it is an isosceles triangle. Therefore, its base angles are equal. Let $\angle OAB = x$, which means $\angle OBA = x$. Similarly, triangle $OAC$ is isosceles ($OA = OC$), so let $\angle OAC = y$ and $\angle OCA = y$.

Step 4: Apply the exterior angle theorem

The exterior angle theorem states that the outside angle of a triangle equals the sum of the two opposite inside angles. For triangle $OAB$, the exterior angle is $\angle BOD$. Therefore, $\angle BOD = x + x = 2x$. For triangle $OAC$, the exterior angle is $\angle COD$. Therefore, $\angle COD = y + y = 2y$.

Step 5: Combine to find the final angles

The total angle at the circumference ($\angle BAC$) is simply $x + y$. The total angle at the centre ($\angle BOC$) is $2x + 2y$, which factorises to $2(x + y)$. Therefore, the angle at the centre is exactly twice the angle at the circumference.

Working with Calculations

When a question asks you to "Apply" this theorem, you will need to calculate missing angles using numerical values or algebraic expressions.

A circle with centre $O$ has points $A$, $B$, and $C$ on its circumference. The angle at the centre, $\angle AOB$, is $146^\circ$. Calculate the angle at the circumference, $\angle ACB$, subtended by the same arc.

Step 1: Identify the relationship.

• $\text{Angle at centre} = 2 \times \text{Angle at circumference}$
• $146^\circ = 2 \times \angle ACB$

Step 2: Rearrange and solve.

• $\angle ACB = 146^\circ \div 2$

Step 3: State the final answer.

• $\angle ACB = 73^\circ$

Algebraic Substitution

Examiners often test this theorem by replacing the angles with algebraic expressions in the same segment.

In a circle with centre $O$, the inscribed angle is $(3x - 10)^\circ$ and the central angle is $160^\circ$. Both angles are subtended by the same arc. Find the value of $x$.

Step 1: Set up the equation using the 2:1 rule.

• $160 = 2 \times (3x - 10)$

Step 2: Expand the brackets.

• $160 = 6x - 20$

Step 3: Rearrange to isolate $x$.

• $180 = 6x$

Step 4: Solve for $x$.

• $x = 30$