1. Chapter at a Glance
How can we find the height of a tall tree or a temple tower without climbing it? Measure its shadow, measure the shadow of a stick at the same moment, and use similar triangles. Two figures are similar when they have the same shape but possibly different sizes — like a photograph and its enlargement. This chapter develops the Basic Proportionality Theorem (Thales' theorem) and the criteria for similarity of triangles, which are used throughout geometry and trigonometry. The BPT proof is one of the most frequently asked board-exam proofs.
- Similar figures and similar polygons.
- Basic Proportionality Theorem (BPT) — statement and proof.
- Converse of BPT.
- Criteria for similarity: AAA (AA), SSS, SAS.
- Using similarity to find lengths and heights.
2. Similar Figures
- All circles are similar; all squares are similar; all equilateral triangles are similar.
- Congruent figures are similar, but similar figures need not be congruent.
Two polygons with the same number of sides are similar if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (proportional). Both conditions are needed for general polygons — a square and a rectangle have equal angles but are not similar; a square and a rhombus have proportional sides but are not similar.
3. Basic Proportionality Theorem (Thales)
If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
In △ABC, if DE ∥ BC (D on AB, E on AC), then AD/DB = AE/EC.
- Join BE and CD. Draw DM ⊥ AC and EN ⊥ AB.
- ar(ADE) = ½ × AD × EN and ar(BDE) = ½ × DB × EN. So ar(ADE)/ar(BDE) = AD/DB. … (1)
- Similarly, ar(ADE) = ½ × AE × DM and ar(DEC) = ½ × EC × DM. So ar(ADE)/ar(DEC) = AE/EC. … (2)
- △BDE and △DEC are on the same base DE and between the same parallels DE and BC, so ar(BDE) = ar(DEC). … (3)
- From (1), (2), (3): AD/DB = AE/EC. ∎
Useful forms: AD/AB = AE/AC and DB/AB = EC/AC also hold.
If a line divides two sides of a triangle in the same ratio, it is parallel to the third side.
In △ABC, DE ∥ BC, AD = 1.5 cm, DB = 3 cm and AE = 1 cm. Find EC.
Solution: 1.5/3 = 1/EC → EC = 2 cm.
4. Criteria for Similarity of Triangles
For triangles we do not need to check both angles and sides — any one of these criteria is enough. We write △ABC ~ △DEF (order of letters shows correspondence).
| Criterion | Condition | Conclusion |
|---|---|---|
| AAA | ∠A = ∠D, ∠B = ∠E, ∠C = ∠F | △ABC ~ △DEF |
| AA | Two pairs of angles equal (third follows) | Similar |
| SSS | AB/DE = BC/EF = CA/FD | Similar |
| SAS | One angle equal and the sides including it proportional | Similar |
When triangles are similar, corresponding sides are proportional and corresponding angles are equal.
A 6 m pole casts a 4 m shadow. At the same time a tower casts a 28 m shadow. Find the height of the tower.
Solution: The sun's rays make the same angle, and both pole and tower are vertical, so the triangles are similar (AA). h/6 = 28/4 → h = 42 m.
A girl 90 cm tall walks away from a 3.6 m lamp post at 1.2 m/s. Find the length of her shadow after 4 seconds.
Solution: She is 4.8 m away. Let the shadow be x: 3.6/(4.8 + x) = 0.9/x → 3.6x = 4.32 + 0.9x → x = 1.6 m.
5. Enrichment: Ratios in Similar Triangles
If △ABC ~ △DEF with ratio of sides k, then perimeters are in ratio k, and corresponding altitudes, medians and angle bisectors are in ratio k. (The ratio of their areas is k², a result you may meet later.)
6. Solved Problems (Board Pattern)
ABCD is a trapezium with AB ∥ DC, and its diagonals meet at O. Show that AO/BO = CO/DO.
Solution: In △AOB and △COD: ∠OAB = ∠OCD and ∠OBA = ∠ODC (alternate angles, AB ∥ DC). So △AOB ~ △COD (AA), giving AO/CO = BO/DO, i.e. AO/BO = CO/DO. ∎
In △ABC, DE ∥ BC with AD = x, DB = x − 2, AE = x + 2, EC = x − 1. Find x.
Solution: x/(x − 2) = (x + 2)/(x − 1) → x² − x = x² − 4 → x = 4.
In right △ABC (∠B = 90°), BD ⊥ AC. Show that △ADB ~ △ABC and BD² = AD × DC.
Solution: △ADB and △ABC share ∠A and both have a right angle → similar (AA). Similarly △BDC ~ △ABC, so △ADB ~ △BDC → AD/BD = BD/DC → BD² = AD × DC. ∎
7. Master Formula Table
| Result | Statement |
|---|---|
| BPT | DE ∥ BC ⇒ AD/DB = AE/EC |
| BPT (whole side) | AD/AB = AE/AC |
| Converse BPT | AD/DB = AE/EC ⇒ DE ∥ BC |
| Similar triangles | AB/DE = BC/EF = CA/FD |
| Perimeter ratio | = ratio of sides |
| Altitude on hypotenuse | BD² = AD × DC |
8. Master Summary Table
| Idea | Key condition | Exam focus |
|---|---|---|
| Similar polygons | Equal angles AND proportional sides | Counterexamples |
| BPT | Line ∥ one side | Proof (3 marks); find lengths |
| Converse BPT | Equal ratios | Prove lines parallel |
| AA / SSS / SAS | Choose the fitting criterion | Prove similarity, then use ratios |
| Applications | Shadows, lamp posts, mirrors | Heights and distances |
9. Mind Map
- Same shape
- Angles + ratios
- ∥ line → equal ratios
- Converse
- AA
- SSS, SAS
- Shadow heights
- Trapezium, right triangles
10. Tips and Tricks
- Write the similarity statement with matching vertex order before writing ratios.
- Parallel lines give equal alternate/corresponding angles → AA similarity.
- Draw the two triangles separately if they overlap in the figure.
- Mixing AD/DB with AD/AB in BPT.
- Using SAS similarity when the equal angle is not between the proportional sides.
- Assuming similar means congruent.
11. Important One-liners
- All congruent figures are similar, but not conversely.
- All equilateral triangles are similar.
- A line parallel to one side divides the other two sides proportionally.
- Two angles equal is enough for triangles to be similar.
- Similar triangles have proportional corresponding sides.
12. Exam-pattern Questions with Detailed Answers
Which pair is always similar?
- Two rectangles
- Two rhombuses
- Two squares
- Two isosceles triangles
Show answer
Squares always have equal angles and proportional sides.
Answer: (C)
In △ABC, DE ∥ BC, AD/DB = 3/5 and AE = 4.8 cm. Find EC.
Show answer
4.8/EC = 3/5 → EC = 8 cm.
Answer: 8 cm
△ABC ~ △PQR with AB = 4, PQ = 6, and perimeter of ABC = 18 cm. Find the perimeter of PQR.
Show answer
Ratio 4 : 6 = 2 : 3 → 18 × 3/2 = 27 cm.
Answer: 27 cm
Points E and F on PQ and PR of △PQR have PE = 4, EQ = 4.5, PF = 8, FR = 9 (cm). Is EF ∥ QR?
Show answer
PE/EQ = 4/4.5 = 8/9; PF/FR = 8/9. Equal → EF ∥ QR (converse of BPT).
Answer: Yes
A vertical stick 20 m long casts a shadow 10 m long. At the same time a tower casts a 50 m shadow. Find the tower's height.
Show answer
h/20 = 50/10 → h = 100 m.
Answer: 100 m
In △ABC, D and E are on AB and AC with DE ∥ BC. If AD = 2 cm, AB = 6 cm and BC = 9 cm, find DE.
Show answer
△ADE ~ △ABC (AA): DE/BC = AD/AB = 1/3 → DE = 3 cm.
Answer: 3 cm
ABCD is a trapezium with AB ∥ DC and diagonals meeting at O. If AO = 3x − 1, OC = 5x − 3, BO = 2x + 1, OD = 6x − 5, find x.
Show answer
AO/OC = BO/OD → (3x − 1)(6x − 5) = (2x + 1)(5x − 3) → 18x² − 21x + 5 = 10x² − x − 3 → 8x² − 20x + 8 = 0 → 2x² − 5x + 2 = 0 → x = 2 or ½. x = ½ makes OC = −0.5 (impossible), so x = 2.
Answer: x = 2
Sides of two triangles are 3, 4, 6 and 9, 12, 18. Are they similar? Which criterion?
Show answer
3/9 = 4/12 = 6/18 = 1/3 → similar by SSS.
Answer: Yes, SSS
Prove that the line drawn through the midpoint of one side of a triangle, parallel to another side, bisects the third side (using BPT).
Show answer
Let D be the midpoint of AB and DE ∥ BC. By BPT, AD/DB = AE/EC. Since AD = DB, AD/DB = 1, so AE = EC. ∎
Answer: Proof as shown
In △ABC, AD is the altitude to BC and AD² = BD × DC. Prove that ∠BAC = 90°.
Show answer
AD/BD = DC/AD and ∠ADB = ∠CDA = 90°, so △ADB ~ △CDA (SAS). Hence ∠BAD = ∠ACD. In △ADC, ∠DAC + ∠ACD = 90°, so ∠DAC + ∠BAD = 90°, i.e. ∠BAC = 90°. ∎
Answer: Proof as shown
If AD and PM are medians of △ABC and △PQR respectively and △ABC ~ △PQR, prove that AB/PQ = AD/PM.
Show answer
AB/PQ = BC/QR = (2BD)/(2QM) = BD/QM and ∠B = ∠Q. So △ABD ~ △PQM (SAS) → AB/PQ = AD/PM. ∎
Answer: Proof as shown
In △ABC, D is on BC such that ∠ADC = ∠BAC. Show that CA² = CB × CD.
Show answer
In △ADC and △BAC: ∠ADC = ∠BAC (given), ∠C common → △ADC ~ △BAC (AA). So CA/CB = CD/CA → CA² = CB × CD. ∎
Answer: Proof as shown
13. Quick Revision Summary
- Similar figures: same shape; polygons need equal angles and proportional sides.
- BPT: DE ∥ BC ⇒ AD/DB = AE/EC (proof using areas); converse is true.
- Triangle similarity: AA, SSS, SAS.
- Similar triangles: corresponding sides and perimeters in the same ratio.
- Use for heights (shadows), trapezium diagonals, right-triangle altitudes.