Circle Geometry Riders

A rider is a geometry problem that combines several theorems in a new diagram. Riders carry many of the Euclidean geometry marks in Paper 2, and every statement needs a correct reason to earn its mark.

Multi-step riders combining the circle theorems

A good routine for any rider:

Mark the diagram. Fill in every given angle, length, tangent, diameter and parallel line.
Look for the theorems. Radii mean isosceles triangles. A diameter means a right angle. A tangent means 90° to the radius and the tan-chord theorem. Four points on a circle mean a cyclic quadrilateral.
Work towards what is asked. If you need to prove two sides equal, look for two equal angles opposite them.
Use earlier parts. "Hence" means the previous answer is needed.

Example. $O$ is the centre and $\hat{B O C} = 140°$ , with $A$ on the major arc. Then $\hat{B A C} = 70°$ (angle at centre), $\hat{O C B} = 20°$ (isosceles triangle $O B C$ ), and if $C T$ is the tangent at $C$ , $\hat{B C T} = 70°$ (tan chord theorem).

Writing statements with reasons

Write each statement on its own line with its reason in brackets:

$\hat{A} = 70°$ ( $∠$ at centre $= 2 \times ∠$ at circumference)
$O B = O C$ (radii)
$\hat{B}_{1} = \hat{C}_{1}$ ( $∠$ s opp equal sides)

Standard reasons to learn:

$∠$ s in the same seg; $∠$ in semi-circle
opp $∠$ s of cyclic quad; ext $∠$ of cyclic quad
tan $⊥$ radius; tans from common pt; tan chord theorem
line from centre $⊥$ chord; line from centre to midpt of chord
alt $∠$ s; corresp $∠$ s; co-int $∠$ s ( $A B ∥ C D$ )
sum of $∠$ s in $△$ ; ext $∠$ of $△$ ; $∠$ s on a str line; vert opp $∠$ s

Every statement needs a correct reason. A correct angle with no reason usually earns nothing.

Key terms

Rider: A problem that applies known theorems to a new diagram.
Statement: A fact you claim, such as an angle size or equal lengths.
Reason: The theorem or definition that justifies the statement.

Common mistakes

Writing statements without reasons.
Using properties from the diagram that are not given, such as a line looking like a diameter.
Using a cyclic quadrilateral property before proving it is cyclic.
Mixing up 'alt $∠$ s' and 'corresp $∠$ s' for parallel lines.

Practice questions

Reveal steps one at a time, then tick the marks you earned.

Question 1

4 marks total

routine

$O$ is the centre of a circle. $A$ , $B$ and $C$ lie on the circle with $A$ on the major arc $B C$ . $\hat{B O C} = 140°$ , and $C T$ is a tangent at $C$ , with $T$ on the opposite side of $B C$ from $A$ .

Calculate, with reasons, $\hat{B A C}$ , $\hat{O C B}$ and $\hat{B C T}$ .

Worked solution

Hidden - try the question first

Solution

$\hat{B A C} = 70°$ ( $∠$ at centre $= 2 \times ∠$ at circumference)
$O B = O C$ (radii), so $\hat{O C B} = \frac{1}{2} (180° - 140°)$
$= 20°$
$\hat{B C T} = \hat{B A C} = 70°$ (tan chord theorem)

Question 2

5 marks total

routine

$A B$ is a diameter of a circle. $C$ and $D$ lie on the circle on the same side of $A B$ , in the order $A$ , $B$ , $D$ , $C$ around the circle. $\hat{C A B} = 25°$ .

Calculate, with reasons:

(a) $\hat{A C B}$ (1)

(b) $\hat{A B C}$ (2)

Worked solution

Hidden - try the question first

Solution

(a) $90°$ ( $∠$ in semi-circle)
(b) $180° - 90° - 25°$ (sum of $∠$ s in $△$ )
$= 65°$
(c) $A B D C$ is a cyclic quad, so $\hat{B D C} + \hat{C A B} = 180°$ (opp $∠$ s of cyclic quad)
$= 20°$

Question 3

6 marks total

complex

In the diagram, $O$ is the centre of the circle, and $A$ , $B$ and $C$ lie on the circle with $C$ on the major arc $A B$ .

Prove that $\hat{O A B} = 90° - \hat{A C B}$ .

Worked solution

Hidden - try the question first

Solution

$\hat{B A C} = 70°$ ( $△$ at centre $= 2 \times ∠$ at circumference)
$O A = O B$ (radii)
$\hat{O A B} = \hat{O B A}$ ( $∠$ s opp equal sides)
$2 \hat{O A B} = 180° - 2 \hat{A C B}$ (sum of $∠$ s in $△$ )
$\hat{O A B} = 90° - \hat{A C B}$

Question 4

9 marks total

problem-solving

In the diagram, $A B C D$ is a cyclic quadrilateral. The tangent $T A S$ at $A$ is parallel to the diagonal $B D$ , with $T$ on the same side as $B$ .

(a) Prove that $A B = A D$ . (5)

(b) Prove that $C A$ bisects $\hat{B C D}$ . (4)

Worked solution

Hidden - try the question first

Solution

(a) $90°$ (alt $∠$ s; $T S ∥ B D$ )
$\hat{T A B} = \hat{A D B}$ (tan chord theorem)
So $\hat{A B D} = \hat{A D B}$
In $△ A B D$ , the sides opposite these angles are $A D$ and $A B$
$A B = A D$ (sides opp equal $∠$ s)
(b) $\hat{A C B} = \hat{A D B}$ ( $∠$ s in the same seg)
$\hat{A C D} = \hat{A B D}$ ( $∠$ s in the same seg)
$\hat{A D B} = \hat{A B D}$ (from (a))
So $\hat{A C B} = \hat{A C D}$ and $C A$ bisects $\hat{B C D}$

FAQs

How do I get started on a geometry rider?

Mark everything you are given on the diagram, then look for radii, diameters, tangents and cyclic quadrilaterals. Each one points to a theorem.

Do I lose marks for missing reasons?

Usually, yes. Geometry marks are normally given for a statement together with its reason, so a missing or wrong reason costs marks.

How are riders marked in the IEB exam?

Usually one mark for each correct statement with its reason, so even a partly finished rider can earn marks.

Multi-step riders combining the circle theorems

Writing statements with reasons

Key terms

Common mistakes

Practice questions

FAQs

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