Try a short problem first. Then use the topic map to plan what to learn next.
Jump to the full topic map · Continue the four-subject practice plan
Worked example: the midpoint of an isosceles triangle
In triangle ABC, suppose AB = AC and M is the midpoint of BC. Prove that AM is perpendicular to BC. Prerequisites: side-side-side congruence and angles on a straight line.
Compare triangles ABM and ACM. We have AB = AC by assumption, BM = CM because M is the midpoint, and AM = AM because this side is shared. The triangles are congruent by SSS.
Corresponding angles AMB and AMC are therefore equal. Since B, M and C are collinear, those two adjacent angles add to 180°. Each is 90°, so AM is perpendicular to BC.
A sketch alone is not proof: the right angles follow from congruence and collinearity. The equal-side and midpoint hypotheses must both appear in the argument.
Continue the same configuration
- Prove that AM bisects angle BAC.
- If AB = 13 and BC = 10, find AM.
Solutions
1. The same congruence gives angle BAM = angle MAC, which is the definition of angle bisection.
2. BM = BC/2 = 5. We have already proved that angle AMB is right. Pythagoras gives AM² = AB² − BM² = 169 − 25 = 144. A length is positive, so AM = 12.
These are teaching examples written for this guide, not past-paper questions or an official marking scheme.
Topic map and learning goals
| Unit | Core content | Advanced or extension content |
|---|---|---|
| Angles and lines | Parallel lines, angle chasing, perpendicularity | Directed angles, particularly for circle configurations |
| Triangles | Congruence, similarity, angle bisectors, medians, altitudes, area ratios | Carefully chosen auxiliary triangles and ratio chains |
| Triangle centres | Circumcentre, incentre, orthocentre, centroid | Excentres, Euler line and nine-point circle |
| Circles | Chords, tangents, inscribed angles, cyclic quadrilaterals | Power of a point, radical axes and intersecting-circle configurations |
| Concurrency and collinearity | Ceva and Menelaus, ratio methods | Trigonometric forms and projective ideas as selective extensions |
| Metric geometry | Pythagoras, sine rule, cosine rule, triangle area formulas | Stewart, Ptolemy and identities involving inradius and circumradius |
| Transformations | Reflection, rotation, translation, homothety | Spiral similarity and inversion |
| Analytic methods | Coordinates, straight lines and circles, vectors | Complex coordinates and barycentric methods where useful |
| Construction | Standard ruler-and-compass constructions, existence | Prove that the constructed object satisfies all requirements |
| Geometric combinatorics | Points, lines, polygons, area, convexity | Extremal configurations, covering and packing arguments |
Learning outcomes: recognise an applicable theorem, introduce a useful auxiliary object, and prove the target relation. A diagram helps discovery; measured lengths and angles do not replace proof.
Working with diagrams: draw and label the configuration. Add a second diagram when introducing an auxiliary circle, projection or transformed point. Separate given information from what still needs proof.
Suggested order: angles → congruence/similarity → circles → ratios and centres → concurrency → transformations → selective analytic methods. Advanced machinery should follow, rather than conceal, geometric understanding.