Lesson 5: Triangles and Congruence

NCERT's Triangles chapter has one central idea: congruence tells us when two triangles are identical, and similarity tells us when they have the same shape at possibly different sizes. Similarity is the more powerful tool — it is how we measure heights we cannot climb and distances we cannot cross. This lesson covers the congruence rules, the similarity criteria, the Basic Proportionality Theorem, and the areas-of-similar-triangles theorem, in NCERT order.

Congruence: SAS, ASA, SSS, RHS

Two triangles are congruent if one can be placed exactly on the other. The tests:

• SAS: two sides and the *included* angle equal.
• ASA: two angles and the included side equal (AAS follows, since angles sum to 180°).
• SSS: all three sides equal.
• RHS: right angle, hypotenuse, and one side equal (for right triangles only).

There is no SSA test: two sides and a non-included angle can produce two different triangles. Remembering what *fails* is as valuable as remembering what works.

Similarity criteria

Two triangles are similar (written △ABC ~ △DEF) if their corresponding angles are equal and corresponding sides are proportional. The working tests mirror congruence:

• AA: two angles equal (the third follows).
• SSS similarity: all three pairs of sides in the same ratio.
• SAS similarity: two pairs of sides proportional and the included angle equal.

Basic Proportionality Theorem (Thales)

If a line parallel to one side of a triangle cuts the other two sides, it divides them in the same ratio: in △ABC with DE ∥ BC, AD/DB = AE/EC. The converse also holds.

Worked example 1 (basic). In △ABC, DE ∥ BC with AD = 3 cm, DB = 6 cm, AE = 2 cm. Find EC. AD/DB = AE/EC gives 3/6 = 2/EC, so EC = 2 × 6/3 = 4 cm.

Worked example 2 (intermediate — areas of similar triangles). The areas of two similar triangles are in the ratio of the *squares* of corresponding sides. If △ABC ~ △DEF with BC/EF = 2/3 and ar(ABC) = 36 cm², then ar(ABC)/ar(DEF) = (2/3)² = 4/9, so ar(DEF) = 36 × 9/4 = 81 cm².

Worked example 3 (NCERT exemplar style). In △ABC, D and E lie on AB and AC with AD = 2.4 cm, AE = 3.2 cm and EC = 4.8 cm. If DE ∥ BC, find DB. AD/DB = AE/EC gives 2.4/DB = 3.2/4.8 = 2/3, so DB = 2.4 × 3/2 = 3.6 cm. Notice the pattern: exemplar problems hand you three of the four lengths and the parallel condition — BPT does the rest in one line.

Pythagoras and its converse

In a right triangle, (hypotenuse)² = (leg)² + (leg)²; conversely, if a² + b² = c² for the sides of a triangle, the angle opposite c is a right angle. The triples worth recognising instantly: 3-4-5, 6-8-10, 5-12-13, 9-12-15, 7-24-25, 8-15-17. A ladder/football-field diagonal problem is almost always a hidden triple.

Competency application: measuring a tower by its shadow

This is the classical Indian surveyor's trick and appears in board papers in modern dress. A vertical stick of height 1.5 m casts a shadow of 2 m at the same moment a tower casts a shadow of 24 m. The Sun's rays are parallel, so the triangles (stick, its shadow) and (tower, its shadow) are similar by AA. Hence tower/1.5 = 24/2 = 12, and the tower is 18 m tall. The assumptions matter and earn marks when stated: same time of day (same angle of elevation), both objects vertical, ground flat.

Common mistakes

• Matching vertices in the wrong order. Writing △ABC ~ △DEF asserts A↔D, B↔E, C↔F. All ratios must use *corresponding* sides; a sloppy statement silently scrambles them.
• Using the area ratio as the side ratio. Areas scale with the *square*. If sides are in ratio 2:3, areas are 4:9 — Worked example 2 is the standard trap in reverse.
• Applying BPT without the parallel condition. AD/DB = AE/EC is only valid when DE ∥ BC (or when the converse is being proved).
• Claiming congruence by SSA. Not a valid test — ever.

Exam Experience

Section A — Multiple choice (1 mark each)

Q1. △ABC ~ △DEF with BC/EF = 2/3. If ar(ABC) = 36 cm², then ar(DEF) is: (a) 54 cm² (b) 81 cm² (c) 24 cm² (d) 144 cm². Answer: (b) 81 cm². Areas are in ratio 4:9, so ar(DEF) = 36 × 9/4 = 81.

Q2. Which of the following is NOT a valid congruence test? (a) SAS (b) ASA (c) SSA (d) RHS. Answer: (c) SSA. Two sides and a non-included angle do not determine a unique triangle.

Section B — Short answer (3 marks) Q3. In △ABC, DE ∥ BC. If AD = 4 cm, AB = 12 cm and AC = 18 cm, find AE. Marking scheme: 1 mark for citing the Basic Proportionality Theorem; 1 mark for AD/AB = AE/AC, i.e. 4/12 = AE/18 (note: AD/DB form with DB = 8 is equally acceptable); 1 mark for AE = 6 cm.

Section C — Case-based long answer (5 marks) Q4. A surveyor wants the width of a river. She stands at point A on one bank facing a tree T on the other bank. She walks 30 m along the bank to point B, where the angle between the bank and the line to the tree is such that the sight lines form similar triangles with a scale model she draws: in her model, the river width is represented by 3 cm and the bank walk by 5 cm; on the ground the bank walk is 30 m. (i) What similarity fact justifies using the model? (1 mark) (ii) Find the actual width of the river. (2 marks) (iii) Later she repeats the method from a bank walk of 45 m and her model gives a width of 4.5 cm for the same 5 cm bank representation. Is this consistent with answer (ii)? Show why. (2 marks) Marking scheme: (i) The sight-line triangles share the same angles (AA similarity), so ratios of corresponding sides are equal (1 mark). (ii) Scale: 5 cm ↔ 30 m, so 1 cm ↔ 6 m (1 mark); width = 3 × 6 = 18 m (1 mark). (iii) New scale: 5 cm ↔ 45 m means 1 cm ↔ 9 m, width = 4.5 × 9 = 40.5 m ≠ 18 m — inconsistent; a consistent model on the new scale would show 18/9 = 2 cm for the width (1 mark for computing, 1 mark for the conclusion/expected value).

Going deeper: the altitude-on-hypotenuse theorem

Drop a perpendicular from the right angle of △ABC (right-angled at A) to the hypotenuse BC, meeting it at D. Then △ADB ~ △ABC and △ADC ~ △ABC (AA: shared acute angle plus the right angle). From these similarities flow the geometric-mean relations: BD × BC = AB², CD × BC = AC², and AD² = BD × DC. These prove the Pythagoras theorem itself: AB² + AC² = BC(BD + DC) = BC × BC = BC². NCERT's proof of Pythagoras uses exactly this similarity route — a similarity argument proving the most famous theorem in mathematics is worth the two minutes it takes to reproduce.

Going deeper: area ratios as a detection tool

If two triangles have equal areas and share the same height, their bases are equal; if they share a base, equal areas force equal heights. The sophisticated exemplar use: DE ∥ BC cuts △ABC so that ar(△ADE) = ar(trapezium DBCE) — the two parts equal, so ar(ADE)/ar(ABC) = 1/2. Since the triangles are similar, (AD/AB)² = 1/2, giving AD/AB = 1/√2. Equal-area conditions almost always become *square-root* ratios of lengths — a signature worth recognising on sight.

Going deeper: medians and the centroid

The three medians of a triangle meet at the centroid, which divides each median in the ratio 2 : 1 (vertex : midpoint). A useful companion is Apollonius's theorem: AB² + AC² = 2(AD² + BD²) where D is the midpoint of BC — Pythagoras applied twice in disguise. Example: in a triangle with AB = 5, AC = 7 and BC = 8, the median to BC has BD = 4, so 25 + 49 = 2(AD² + 16), giving AD² = 37 − 16 = 21, AD = √21 ≈ 4.58. Centroid and median facts pair naturally with Lesson 7's centroid *coordinates* — the same point, described geometrically here and algebraically there.

One more theorem: the medial triangle

Joining the midpoints of a triangle's sides (the mid-point theorem of the main lesson, applied three times) divides it into four congruent triangles, each with area exactly 1/4 of the whole and perimeter exactly 1/2 of the whole. So a triangle with sides 6, 8, 10 and area 24 has medial triangle perimeter 12 and area 6. The mid-point segment is parallel to the third side and half its length — which is also why a line through one midpoint parallel to a side bisects the other side (the converse). Surveyors use the same fact in reverse: an inaccessible side's length can be measured from its accessible mid-segment.

Final drills

Drill 1 — area split by a parallel line. In △ABC, DE ∥ BC with AD : DB = 2 : 1. Find ar(△ADE) : ar(△ABC). The similarity ratio is AD/AB = 2/3, so the area ratio is (2/3)² = 4 : 9. (And the trapezium takes the remaining 5/9 — a common follow-up.) Ratios along a side translate to the *square* for areas: commit the chain length-ratio → area-ratio² to reflex.

Drill 2 — why a rectangle's diagonals are equal (proof skeleton). In rectangle ABCD, compare △ABC and △BAD: AB is common, BC = AD (opposite sides), and ∠ABC = ∠BAD = 90°. By SAS the triangles are congruent, so AC = BD ∎. Most "prove the diagonals equal/bisect" results in this chapter are one well-chosen congruent pair away — when stuck, hunt for two triangles that share a side and bracket the segment you care about.