1. Angles and Angle Properties
Angle properties are the foundation of all geometry. Every geometry problem depends on knowing and applying these rules precisely and in sequence.
Basic Angle Facts
| Rule | Statement | Diagram Note |
|---|---|---|
| Angles on a straight line | Angles on a straight line add up to 180°. | a + b + c = 180° |
| Angles at a point | Angles around a full point add up to 360°. | a + b + c + d = 360° |
| Vertically opposite angles | When two lines cross, opposite angles are equal. | a = c and b = d |
| Angles in a triangle | The three interior angles of any triangle add up to 180°. | a + b + c = 180° |
| Exterior angle of a triangle | An exterior angle equals the sum of the two non-adjacent interior angles. | ext = a + b |
| Angles in a quadrilateral | The four interior angles add up to 360°. | a + b + c + d = 360° |
Parallel Lines — Three Key Angle Pairs
Corresponding Angles
Equal. In the same position at each intersection (F-shape).
a = b (F angles)
Alternate Angles
Equal. On opposite sides of the transversal between the parallel lines (Z-shape).
a = b (Z angles)
Co-interior Angles
Add up to 180°. On the same side of the transversal between the parallel lines (C-shape).
a + b = 180° (C angles)
Polygon Angle Formulae
Interior Angle Sum
where n = number of sides
Each interior angle of a regular polygon = (n−2)×180° ÷ n
Exterior Angle Sum
Each exterior angle of a regular polygon = 360° ÷ n
Interior + Exterior = 180° (angles on a straight line)
📐 Worked Example 1 — Regular Polygon Angles
A regular polygon has an interior angle of 156°. Find the number of sides.
n = 360 ÷ 24 = 15 sides
2. Triangles — Properties, Congruence, and Similarity
Types of Triangles
| Type | Properties |
|---|---|
| Equilateral | All three sides equal. All three angles = 60°. |
| Isosceles | Two sides equal. Base angles (opposite equal sides) are equal. |
| Scalene | All sides and angles different. |
| Right-angled | One angle = 90°. Side opposite 90° is the hypotenuse (longest side). |
| Obtuse-angled | One angle > 90°. |
| Acute-angled | All angles < 90°. |
Congruent Triangles
SSS — Side Side Side
All three sides of one triangle are equal to the three sides of the other.
SAS — Side Angle Side
Two sides and the included angle (between those sides) are equal.
ASA — Angle Side Angle
Two angles and the included side are equal.
RHS — Right Hypotenuse Side
Both triangles have a right angle, equal hypotenuses, and one other equal side.
Similar Triangles
If two triangles are similar with scale factor k (ratio of corresponding sides):
• Ratio of corresponding lengths = k
• Ratio of corresponding areas = k²
• Ratio of corresponding volumes = k³ (for 3D shapes)
📐 Worked Example 2 — Similar Triangles
Triangles ABC and PQR are similar with AB:PQ = 3:5. Area of triangle ABC = 27 cm². Find the area of triangle PQR.
📐 Worked Example 3 — Finding Unknown Sides in Similar Triangles
In the diagram, DE is parallel to BC. AD = 4, DB = 6, DE = 5. Find BC.
3. Circle Theorems
Circle theorems are one of the most heavily tested topics in Cambridge O Level Geometry. You must know all eight theorems, be able to state the reason for each step, and apply multiple theorems in a single problem.
The Eight Circle Theorems
| # | Theorem | Statement |
|---|---|---|
| 1 | Angle at Centre | The angle subtended at the centre by an arc is twice the angle subtended at the circumference by the same arc. Reason to write: "angle at centre = 2 × angle at circumference (same arc)" |
| 2 | Angle in Semicircle | The angle in a semicircle (angle subtended by a diameter at the circumference) is always 90°. Reason: "angle in a semicircle = 90°" |
| 3 | Angles in Same Segment | Angles subtended by the same arc (chord) on the same side are equal. Reason: "angles in the same segment" |
| 4 | Cyclic Quadrilateral | Opposite angles of a cyclic quadrilateral (vertices on the circle) add up to 180°. Reason: "opposite angles of a cyclic quadrilateral" |
| 5 | Tangent-Radius | A tangent to a circle is perpendicular to the radius at the point of tangency. Reason: "tangent ⊥ radius" |
| 6 | Two Tangents from External Point | Two tangents drawn from an external point to a circle are equal in length. The line joining the external point to the centre bisects the angle between the tangents. Reason: "tangents from an external point are equal" |
| 7 | Alternate Segment Theorem | The angle between a tangent and a chord at the point of tangency equals the inscribed angle on the opposite side of the chord (in the alternate segment). Reason: "alternate segment theorem" |
| 8 | Perpendicular from Centre to Chord | The perpendicular from the centre to a chord bisects the chord. Conversely, the perpendicular bisector of a chord passes through the centre. Reason: "perpendicular from centre bisects chord" |
📐 Worked Example 4 — Multiple Circle Theorems
O is the centre of the circle. A, B, C are points on the circle. Angle BAC = 35°. Find angle BOC and angle OBC. Give reasons.
Angle at centre = 2 × angle at circumference (same arc BC).
Angle BOC = 2 × 35° = 70°
Reason: angle at centre is twice angle at circumference
Triangle OBC is isosceles (OB = OC = radii).
Base angles are equal: angle OBC = angle OCB.
2 × angle OBC = 180° − 70° = 110°
Angle OBC = 55°
Reason: base angles of isosceles triangle (OB = OC, radii)
In circle theorem questions, every angle you find must be accompanied by a written reason in brackets. Without reasons, you lose marks even if the angle value is correct. Use the exact phrases shown in the theorem table above.
4. Constructions and Loci
Constructions use only a ruler and compasses (no protractor). You must leave all construction arcs visible in the exam.
Standard Constructions
| Construction | Steps |
|---|---|
| Perpendicular bisector of AB | 1. Open compasses to more than half of AB. 2. Draw arcs above and below from A. 3. Without changing setting, draw arcs from B. 4. Join the two arc intersections — this is the perpendicular bisector. |
| Angle bisector of angle ABC | 1. Draw arc from B cutting both arms at D and E. 2. Draw equal arcs from D and E intersecting at F. 3. Join BF — this bisects angle ABC. |
| Perpendicular from point P to line L | 1. Draw arc from P cutting line L at two points A and B. 2. Construct perpendicular bisector of AB. 3. This perpendicular passes through P. |
| 60° angle | Draw a line. Mark point A. Draw arc from A. Mark where arc crosses line at B. Open same radius, draw arc from B cutting first arc at C. Join AC — angle BAC = 60°. |
| 90° angle at a point on a line | Construct perpendicular bisector at the required point by placing compass point ON the line and using the standard perpendicular bisector method. |
Standard Loci
1. Locus of points at fixed distance r from point P → circle with centre P, radius r
2. Locus of points equidistant from two points A and B → perpendicular bisector of AB
3. Locus of points equidistant from two intersecting lines → angle bisectors of the lines
4. Locus of points at fixed distance d from a line L → two parallel lines, distance d on each side of L
📐 Worked Example 5 — Combined Loci
A is 6 cm from B. A point P must be: (i) less than 4 cm from A, AND (ii) closer to B than to A. Describe the region where P can be.
5. Symmetry
Line Symmetry (Reflective)
A shape has line symmetry if it can be folded along a line so that both halves match exactly. The fold line is the line (axis) of symmetry.
| Shape | Lines of Symmetry |
|---|---|
| Equilateral triangle | 3 |
| Square | 4 |
| Rectangle | 2 |
| Rhombus | 2 |
| Regular hexagon | 6 |
| Circle | Infinite |
| Isosceles triangle | 1 |
| Parallelogram | 0 |
Rotational Symmetry
A shape has rotational symmetry if it maps onto itself when rotated by less than 360° about its centre. The order is how many times it maps onto itself in a full 360° turn.
| Shape | Order |
|---|---|
| Equilateral triangle | 3 |
| Square | 4 |
| Rectangle | 2 |
| Rhombus | 2 |
| Regular hexagon | 6 |
| Circle | Infinite |
| Parallelogram | 2 |
| Scalene triangle | 1 (none) |
• Number of lines of symmetry = n
• Order of rotational symmetry = n
A shape with rotational symmetry of order 1 has NO rotational symmetry (it only maps onto itself after a full 360° turn).
6. Properties of Quadrilaterals
| Shape | Sides | Angles | Diagonals | Symmetry |
|---|---|---|---|---|
| Square | All 4 equal, all parallel in pairs | All 90° | Equal, perpendicular, bisect each other | 4 lines, order 4 |
| Rectangle | Opposite sides equal and parallel | All 90° | Equal, bisect each other (not perpendicular) | 2 lines, order 2 |
| Rhombus | All 4 equal, opposite sides parallel | Opposite angles equal | Perpendicular, bisect each other (not equal) | 2 lines, order 2 |
| Parallelogram | Opposite sides equal and parallel | Opposite angles equal | Bisect each other (not equal, not perpendicular) | 0 lines, order 2 |
| Trapezium | One pair of parallel sides | Co-interior angles between parallel sides add to 180° | No special properties | 0 lines (isosceles trapezium: 1 line) |
| Kite | Two pairs of adjacent equal sides | One pair of equal opposite angles | One diagonal perpendicular bisector of the other | 1 line, order 1 |
📝 Exam Practice Questions
Q1 [3 marks] — A regular polygon has 12 sides. Find (a) the sum of interior angles (b) each interior angle (c) each exterior angle.
(b) Each interior = 1800° ÷ 12 = 150°
(c) Each exterior = 360° ÷ 12 = 30° (or 180°−150°=30°)
Q2 [3 marks] — O is the centre of a circle. Points A, B and C lie on the circle. Angle ACB = 48°. Find angle AOB and angle OAB. Give reasons for each answer.
Reason: angle at centre = twice angle at circumference (same arc AB)
Angle OAB: Triangle OAB is isosceles (OA = OB = radii).
Angle OAB = (180° − 96°)/2 = 84°/2 = 42°
Reason: base angles of isosceles triangle are equal (OA = OB, both radii)
Q3 [3 marks] — ABCD is a cyclic quadrilateral. Angle DAB = 3x + 10 and angle BCD = 5x − 30. Find the value of x and both angles.
Opposite angles of cyclic quadrilateral add to 180°:
(3x+10) + (5x−30) = 180°
8x − 20 = 180 → 8x = 200 → x = 25
Angle DAB = 3(25)+10 = 85°
Angle BCD = 5(25)−30 = 95°
Check: 85° + 95° = 180° ✓
Q4 [4 marks] — Triangle PQR is similar to triangle XYZ. PQ = 6 cm, QR = 8 cm, PR = 10 cm, XY = 9 cm. Find: (a) YZ and XZ (b) the ratio of the areas of the two triangles.
(a) YZ = QR × k = 8 × 3/2 = 12 cm
XZ = PR × k = 10 × 3/2 = 15 cm
(b) Ratio of areas = k² = (3/2)² = 9:4 (XYZ : PQR)
Q5 [2 marks] — TA and TB are tangents from external point T to a circle with centre O. Angle ATB = 64°. Find angle AOB.
OA ⊥ TA and OB ⊥ TB (tangent ⊥ radius → each = 90°)
Angles in quadrilateral OATB: 90° + 90° + 64° + angle AOB = 360°
Angle AOB = 360° − 244° = 116°
Q6 [3 marks] — Two parallel lines are 4 cm apart. A point P lies between them, 1.5 cm from the first line. A point Q must be equidistant from both parallel lines. Describe the locus of Q and find the distance between P and the locus of Q.
Distance from P to locus of Q:
P is 1.5 cm from line 1, so P is 2.5 cm from line 2.
Locus of Q is 2 cm from line 1.
Distance PQ = 2 − 1.5 = 0.5 cm
Q7 [4 marks] — In triangle ABC, DE is parallel to BC where D is on AB and E is on AC. AD = 3 cm, DB = 5 cm and DE = 4.5 cm. Calculate BC. If the area of triangle ADE = 13.5 cm², find the area of trapezium BCED.
BC = DE × (8/3) = 4.5 × 8/3 = 12 cm
Area of triangle ABC: = Area ADE × k² = 13.5 × (8/3)² = 13.5 × 64/9 = 96 cm²
Area of trapezium BCED = 96 − 13.5 = 82.5 cm²