Lesson 9: Plane Geometry

Geometry is where Chinese mathematics met its most rigorous tradition: the Mohist canon (墨经) defined points, lines and circles with philosophical precision, and Liu Hui's commentary on the Nine Chapters proved results by ingenious dissection — cutting shapes apart and reassembling them, "using the excess to fill the deficit". Modern Gaokao geometry demands formal proof: statements justified step by step from definitions, axioms and earlier theorems. This lesson builds that craft — congruence, similarity, the Pythagorean theorem and its converse, and area on a lattice — with every proof complete. An unjustified claim in a proof earns nothing; the chain of reasons is the answer.

Congruence and similarity

Two triangles are congruent if one can be moved onto the other exactly; the criteria are SSS, SAS, ASA and AAS (note: SSA is not a criterion — two sides and a non-included angle can fit two different triangles). Corresponding vertices must be named in order: △ABC ≅ △DEF asserts A↔D, B↔E, C↔F, and every subsequent claim about sides or angles inherits that correspondence.

Similar triangles have equal angles and proportional sides with ratio k. Areas then scale with k², volumes with k³ (Lesson 4's warning in geometric form). The workhorse configuration: a line parallel to one side of a triangle cuts off a smaller, similar triangle — the basis of most length computations. The Pythagorean theorem a² + b² = c² holds exactly for right triangles, and its converse also holds: if a² + b² = c², the triangle is right-angled. Comparing a² + b² with c² classifies any triangle: greater means acute (opposite c), equal right, less obtuse.

Worked examples

Worked example 1 — base angles of an isosceles triangle. Claim: in △ABC with AB = AC, the base angles satisfy ∠B = ∠C. Proof: let M be the midpoint of BC, and draw AM. In △ABM and △ACM: AB = AC (given), BM = CM (M is the midpoint), AM = AM (common). By SSS the triangles are congruent, so the corresponding angles at B and C are equal. ∎ The construction (one auxiliary line) unlocked everything — choosing the auxiliary line is the creative step in any geometry proof.

Worked example 2 — parallel line similarity. In △ABC, D lies on AB and E on AC with DE ∥ BC. Given AD = 2, DB = 4 and DE = 3, find BC. Since DE ∥ BC, the corresponding angles at D and E equal those at B and C, so △ADE ~ △ABC with ratio AD : AB = 2 : 6 = 1 : 3. Hence BC = 3 × DE = 9. Sanity: the small triangle sits inside the large one, so BC > DE ✓.

Worked example 3 — classifying by Pythagoras. Classify the triangles: (a) sides 7, 24, 25: 7² + 24² = 49 + 576 = 625 = 25² — right-angled. (b) sides 5, 6, 8: longest side 8: 5² + 6² = 61 < 64 = 8² — the angle opposite the side 8 is obtuse, so the triangle is obtuse. The converse does the classifying; never assume a triangle is right-angled because it is drawn that way.

Worked example 4 — parallelogram diagonals. Claim: the diagonals of a parallelogram bisect each other. Proof: let ABCD be a parallelogram with diagonals meeting at O. AB ∥ CD gives alternate angles ∠OAB = ∠OCD and ∠OBA = ∠ODC (parallel lines, transversal = the diagonals). AB = CD (opposite sides of a parallelogram). By ASA, △AOB ≅ △COD. Hence AO = CO and BO = DO — the diagonals bisect each other. ∎

Worked example 5 — Pick's theorem on a lattice. For a lattice polygon, area = I + B/2 − 1, where I counts interior lattice points and B counts boundary points. Apply it to the triangle (0, 0), (4, 0), (0, 3). Boundary points are counted edge by edge: a segment between lattice points contains gcd(|Δx|, |Δy|) lattice steps, so B = gcd(4, 0) + gcd(0, 3) + gcd(4, 3) = 4 + 3 + 1 = 8. (The hypotenuse, from (4, 0) to (0, 3), has gcd(4, 3) = 1: it passes through no lattice point strictly between its endpoints.) Direct computation gives the area: ½ × 4 × 3 = 6. Pick's theorem then forces 6 = I + 8/2 − 1 = I + 3, so I = 3. The gcd rule for slanted edges is the only subtle step; everything else is arithmetic.

Challenge problem — Gaokao standard

Problem. In △ABC, M and N are the midpoints of AB and AC. Prove that MN ∥ BC and MN = BC/2, and find the ratio of the areas of △AMN and △ABC.

Solution. Consider △AMN and △ABC. They share the angle at A. The sides about it: AM/AB = 1/2 and AN/AC = 1/2 (midpoints). By SAS (equal included angle, proportional surrounding sides), △AMN ~ △ABC with ratio 1/2. Therefore MN/BC = 1/2, i.e. MN = BC/2, and the corresponding angles satisfy ∠AMN = ∠ABC; equal corresponding angles for the transversal AB imply MN ∥ BC. Areas scale as the square of the ratio: area(△AMN)/area(△ABC) = (1/2)² = 1/4. This is the midline theorem — the single most useful similarity fact in the curriculum, and the proof shows why similarity criteria are worth memorising exactly.

Common mistakes

• Using SSA as a congruence criterion. It does not exist. Two sides and a non-included angle admit two configurations (the ambiguous case returns in Lesson 12).
• Mismatched correspondence. In △ABC ≅ △DEF, side AB corresponds to DE, not DF. Write congruences with vertices in matching order, always.
• Assuming the diagram. Right angles, midpoints and parallels must be given or proved; a square-looking corner is not a right angle until marked or derived.
• Area ratio equal to length ratio. Similar figures with ratio k have area ratio k². (Challenge problem: ratio 1/2 in length, 1/4 in area.)
• Circular reasoning. Using "base angles are equal" inside the proof that base angles are equal. Audit each proof: does any step secretly assume the conclusion?

Exam Experience

Multiple choice. Two triangles are similar with ratio 2 : 3 (small : large). The small triangle has area 20 cm². The large triangle's area is: A. 30 cm² B. 45 cm² C. 13⅓ cm² D. 40 cm² Answer: B. Areas scale by (3/2)² = 9/4: 20 × 9/4 = 45. Option A applies the length ratio — the designed trap.

Fill in the blank. A triangle has sides 9, 12, 15. Its area is ____. Answer: 54. 9² + 12² = 81 + 144 = 225 = 15², so it is right-angled with legs 9 and 12: area ½ × 9 × 12 = 54.

Solution question (10 marks). In △ABC, AB = AC. Points D and E are the midpoints of AB and AC respectively. Prove that BE = CD. Marking scheme. State the givens algebraically: AD = ½AB and AE = ½AC, and AB = AC gives AD = AE — 2 marks. Identify the two triangles △ABE and △ACD and their shared angle ∠A — 1 mark. In △ABE and △ACD: AB = AC (given) — 1 mark; ∠A = ∠A (common) — 1 mark; AE = AD (proved above) — 1 mark. Conclude △ABE ≅ △ACD by SAS — 2 marks. Therefore the corresponding sides BE = CD — 2 marks. [10 marks] Notice the architecture: every mark is a link in the chain, and the final mark is reserved for the correspondence conclusion. A proof that "sees" the answer but skips links scores half.

Going deeper — the exterior angle and Apollonius

Exterior angle theorem. The exterior angle of a triangle equals the sum of the two remote interior angles. Proof in two lines: at vertex C, the exterior angle is 180° − C (straight line), and A + B + C = 180° gives A + B = 180° − C. Equal. ∎ This modest theorem is an angle-chase accelerator: it converts a two-step chase into one step, and it is the engine behind cyclic-quadrilateral supplements in Lesson 10.

Apollonius's median formula. In a triangle with sides a, b, c, the median to side c has length m satisfying m² = (2a² + 2b² − c²)/4. Worked example: sides 5, 6, 7 (median to the side 7, so set c = 7, a = 5, b = 6): m² = (2 × 25 + 2 × 36 − 49)/4 = (50 + 72 − 49)/4 = 73/4, so m = √73/2 ≈ 4.27. Sanity bounds every median must obey: it is shorter than the average of the adjacent sides ((5 + 6)/2 = 5.5 > 4.27 ✓) and longer than half their difference — quick plausibility screens. The proof is a cosine-rule double application at the midpoint (the two supplementary angles have cosines that cancel) — a forward reference to Lesson 12, and a model of how the curriculum's results interlock. Apollonius also names the configuration behind Lesson 10's power of a point discussions in olympiad training (Lesson 17): the median, the angle bisector and the symmedian are the three classical cevians, and knowing median lengths on sight is competition table stakes.

Second look — the rhombus diagonal proof

Claim. The diagonals of a rhombus are perpendicular. Proof: let ABCD be a rhombus (all sides equal) with diagonals meeting at O. From Worked example 4's theorem (proved there for any parallelogram, and a rhombus is one), the diagonals bisect each other: AO = OC. Consider △AOB and △COB: AB = CB (rhombus sides), AO = CO (just shown), BO = BO (common). By SSS, △AOB ≅ △COB, so ∠AOB = ∠COB. But these two angles form a linear pair along the straight diagonal AC, so they sum to 180°. Equal and supplementary: each is 90°. Hence the diagonals are perpendicular. ∎ Study the architecture: the proof consumes a previous theorem (diagonal bisection) as a lemma, then applies SSS, then a one-line angle argument. Competition proofs are built exactly this way — a tower of small certainties. When you are stuck on a proof, inventory the lemmas you already own about the figure; the missing step is usually an application of one of them, not a new idea.