Triangles — Class 10 Mathematics
1. Similarity of Triangles
Two triangles are SIMILAR if they have the SAME SHAPE (not necessarily the same size). Their corresponding angles are EQUAL and corresponding sides are PROPORTIONAL. Symbol: ΔABC ~ ΔPQR. 'If the ratio of similarity is 1, the triangles are CONGRUENT (~≅). Congruence is a special case of similarity.'
Similarity Criteria
| Criterion | What Must Hold | Comparison with Congruence |
|---|---|---|
| AAA (Angle-Angle-Angle) | ∠A=∠P, ∠B=∠Q, ∠C=∠R | AAA for congruence does NOT exist; for similarity it's VALID |
| SSS (Side-Side-Side) | AB/PQ = BC/QR = CA/RP | In congruence: sides EQUAL. In similarity: sides PROPORTIONAL |
| SAS (Side-Angle-Side) | AB/PQ = AC/PR and ∠A=∠P | In both: one angle + two sides (but proportion vs equality) |
'If two angles of one triangle equal two angles of another, the third angles automatically match (angle sum = 180°). So checking AA is sufficient — this is why the criterion is often written as AA, not AAA.'
2. Basic Proportionality Theorem (Thales Theorem)
Statement: If a line is drawn PARALLEL to one side of a triangle intersecting the other two sides at DISTINCT points, then the other two sides are divided in the SAME RATIO.
In ΔABC, if DE ∥ BC (D on AB, E on AC), then AD/DB = AE/EC.
Proof of Thales Theorem
Given: ΔABC, DE ∥ BC. To Prove: AD/DB = AE/EC. Construction: Join BE and CD. Draw DM ⟂ AC and EN ⟂ AB. Proof: Area of ΔADE = ½×AD×EN. Area of ΔBDE = ½×DB×EN. Ratio: ar(ΔADE)/ar(ΔBDE) = AD/DB ... (1). Similarly, ar(ΔADE)/ar(ΔCDE) = AE/EC ... (2). But ΔBDE and ΔCDE are on the same base DE and between the same parallels DE ∥ BC. So ar(ΔBDE) = ar(ΔCDE) ... (3). From (1), (2), and (3): AD/DB = AE/EC. ✓
Converse of Thales Theorem
If a line divides two sides of a triangle in the SAME RATIO, then the line is PARALLEL to the third side. If AD/DB = AE/EC → DE ∥ BC.
Worked Examples — Thales Theorem
Example 1: In ΔABC, DE ∥ BC. AD = 3 cm, DB = 4 cm, AE = 4.5 cm. Find EC. AD/DB = AE/EC → 3/4 = 4.5/EC → EC = (4.5×4)/3 = 6 cm.
Example 2: In ΔPQR, ST ∥ QR. PS/SQ = 3/5. If PR = 16 cm, find PT and TR. PS/SQ = PT/TR = 3/5. Let PT = 3k, TR = 5k. Total PR = PT+TR = 8k = 16 → k = 2. PT = 6 cm, TR = 10 cm.
Example 3 — Proving the Converse: In ΔABC, D and E are points on AB and AC. AD=2.4cm, DB=3.6cm, AE=3.2cm, EC=4.8cm. Check if DE ∥ BC. AD/DB = 2.4/3.6 = 2/3. AE/EC = 3.2/4.8 = 2/3. Since ratios are equal → DE ∥ BC (converse of BPT).
3. Areas of Similar Triangles
Theorem: The ratio of the AREAS of two similar triangles is equal to the SQUARE of the ratio of their CORRESPONDING SIDES.
If ΔABC ~ ΔPQR, then ar(ΔABC)/ar(ΔPQR) = (AB/PQ)² = (BC/QR)² = (CA/RP)².
Proof Sketch: Express area as ½×base×height. Since triangles are similar, both base AND height scale by the same ratio. So area scales by (ratio)².
Example: ΔABC ~ ΔPQR. AB:PQ = 2:3. ar(ΔABC) = 48 cm². Find ar(ΔPQR). ar(ΔABC)/ar(ΔPQR) = (2/3)² = 4/9. 48/ar(ΔPQR) = 4/9 → ar(ΔPQR) = 48×9/4 = 108 cm².
4. Pythagoras Theorem
Statement: In a RIGHT-ANGLED triangle, the SQUARE of the HYPOTENUSE equals the SUM of the SQUARES of the other two sides. If ∠B = 90° in ΔABC, then AC² = AB² + BC².
Proof Using Similar Triangles
Given: ΔABC, right-angled at B. To Prove: AC² = AB² + BC². Construction: Draw BD ⟂ AC (meets AC at D). Proof: In ΔADB and ΔABC: ∠A = ∠A (common). ∠ADB = ∠ABC = 90°. By AA → ΔADB ~ ΔABC. So AD/AB = AB/AC → AB² = AD×AC ... (1). In ΔBDC and ΔABC: ∠C = ∠C (common). ∠BDC = ∠ABC = 90°. By AA → ΔBDC ~ ΔABC. So CD/BC = BC/AC → BC² = CD×AC ... (2). Add (1) and (2): AB² + BC² = AD×AC + CD×AC = AC(AD+CD) = AC×AC = AC². ✓
Converse of Pythagoras Theorem
If in a triangle, the square of one side equals the sum of squares of the other two, the angle OPPOSITE the first side is a RIGHT ANGLE. If AC² = AB² + BC² → ∠B = 90°.
Applications
Example 1 — Ladder Problem: A ladder 10 m long rests against a wall. Its foot is 6 m from the wall. How high up does the ladder reach? AC=10m, BC=6m. AB² = AC² − BC² = 100−36 = 64. AB = 8 m.
Example 2 — Diagonal: Find the diagonal of a rectangle 15 cm × 8 cm. Diagonal² = 15² + 8² = 225+64 = 289. Diagonal = 17 cm.
Example 3 — Distance: Two poles of heights 6 m and 11 m stand vertically on the ground. The distance between their feet is 12 m. Find the distance between their tops. Height difference = 5 m. Distance between tops = √(12²+5²) = √(144+25) = √169 = 13 m.
Example 4 — Proving a Triangle is Right: Determine if sides 7 cm, 24 cm, 25 cm form a right triangle. Longest = 25. 25² = 625. 7²+24² = 49+576 = 625. Since 25² = 7²+24² → right-angled ✓. (7, 24, 25) is a Pythagorean triplet.
5. Key Theorem Summary
| Theorem | Statement | Use |
|---|---|---|
| Thales (BPT) | DE ∥ BC → AD/DB = AE/EC | Find unknown segments |
| Converse of BPT | AD/DB = AE/EC → DE ∥ BC | Prove lines parallel |
| Areas of similar Δs | ar(Δ₁)/ar(Δ₂) = (side ratio)² | Area problems |
| Pythagoras | ∠B=90° → AC²=AB²+BC² | Find unknown side |
| Converse of Pythagoras | AC²=AB²+BC² → ∠B=90° | Prove right angle |
6. Common Mistakes
- Applying Thales without checking ∥: The theorem only works if the line IS parallel to the third side.
- Using AAA for congruence: AAA proves SIMILARITY, not congruence.
- Hypotenuse misidentification: The hypotenuse is ALWAYS opposite the right angle. It is the longest side.
- Forgetting to SQUARE the ratio for areas: Areas scale as side ratio SQUARED. If sides are 2:3, areas are 4:9, NOT 2:3.
7. AP Exam Focus
| Topic | Marks |
|---|---|
| Thales Theorem (state + prove) | 4-5 |
| Thales Theorem application | 3-4 |
| Pythagoras Theorem proof | 4-5 |
| Pythagoras application problems | 3-4 |
| Areas of similar triangles | 3-4 |
Quick Self-Test
- State Thales Theorem. (Answer: A line parallel to one side divides the other two sides in the same ratio.)
- In ΔABC, DE ∥ BC, AD=4, DB=6, AE=5. Find EC. (Answer: 4/6=5/EC → EC=7.5.)
- Find diagonal of square of side 10 cm. (Answer: √(10²+10²) = √200 = 10√2 ≈ 14.14 cm.)
- ΔLMN ~ ΔPQR, LM:PQ=3:5. ar(ΔLMN)=36cm². Find ar(ΔPQR). (Answer: 36×25/9=100 cm².)
- Is (9, 12, 15) a Pythagorean triplet? (Answer: 15²=225, 9²+12²=81+144=225. Yes.)
