Exercise Solutions Available
Step-by-step answers for all textbook questions in this chapter
Similar Figures
Two figures are similar if they have the same shape but not necessarily the same size.
- All congruent figures are similar, but NOT all similar figures are congruent
- Corresponding angles are equal; corresponding sides are proportional
Basic Proportionality Theorem (BPT / Thales' Theorem)
Statement: If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.
If DE ∥ BC in △ABC, then: $$\frac{AD}{DB} = \frac{AE}{EC}$$
Also: AD/AB = AE/AC = DE/BC
Converse of BPT: If a line divides two sides of a triangle in the same ratio, it is parallel to the third side.
Worked example: In △ABC, DE ∥ BC. If AD = 4 cm, DB = 2 cm, AE = 6 cm, find EC.
By BPT: AD/DB = AE/EC → 4/2 = 6/EC → EC = 3 cm
Criteria for Similarity of Triangles
AA (Angle-Angle) Criterion
If two angles of one triangle are equal to two corresponding angles of another triangle, the triangles are similar.
(Third angles are automatically equal since angles sum to 180°)
SSS (Side-Side-Side) Criterion
If the corresponding sides of two triangles are in the same ratio, the triangles are similar.
$$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP}$$
SAS (Side-Angle-Side) Criterion
If one angle of a triangle equals one angle of another, and the sides including these angles are proportional, the triangles are similar.
Areas of Similar Triangles
Theorem: The ratio of areas of two similar triangles equals the square of the ratio of their corresponding sides.
$$\frac{\text{Area}(\triangle ABC)}{\text{Area}(\triangle PQR)} = \frac{AB^2}{PQ^2} = \frac{BC^2}{QR^2} = \frac{CA^2}{RP^2}$$
Worked example: △ABC ~ △DEF. If AB = 6 cm and DE = 4 cm, find the ratio of their areas.
Ratio of sides = 6:4 = 3:2
Ratio of areas = 3²:2² = 9:4
Pythagoras Theorem
Statement: In a right-angled triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides.
If ∠B = 90° in △ABC: $$AC^2 = AB^2 + BC^2$$
Proof sketch (using similar triangles): Draw BD ⊥ AC. Then △ABC ~ △ADB ~ △BDC. From △ABC ~ △ADB: AB² = AD × AC From △ABC ~ △BDC: BC² = DC × AC Adding: AB² + BC² = AC(AD + DC) = AC × AC = AC²
Converse of Pythagoras Theorem
If AB² + BC² = AC², then ∠B = 90°.
Common Pythagorean triplets:
| Triplet | Relation |
|---|---|
| 3, 4, 5 | 9+16=25 |
| 5, 12, 13 | 25+144=169 |
| 8, 15, 17 | 64+225=289 |
| 7, 24, 25 | 49+576=625 |
Multiples also work: 6,8,10 or 9,12,15 etc.
Important Theorems for Proofs
Theorem: If altitudes of two similar triangles are in ratio k, areas are in ratio k²
Altitude ratio = side ratio, so area ratio = (side ratio)² = k²
Theorem on median
In △ABC, if AD is the median (D is midpoint of BC): AB² + AC² = 2(AD² + BD²) (Apollonius theorem)
Worked Proof-Type Questions
Q. In △ABC, AD is the altitude from A to BC. Prove that AB² + CD² = AC² + BD².
Since AD ⊥ BC: In △ADB: AB² = AD² + BD² ... (i) In △ADC: AC² = AD² + CD² ... (ii) Subtract (ii) from (i): AB² − AC² = BD² − CD² ∴ AB² + CD² = AC² + BD² ✓
Board Exam Tips
- 8–10 marks from this chapter — highest in Geometry section
- BPT and its converse are always in exams — memorise statement exactly
- AA similarity is most commonly used in proofs — look for equal angles first
- Area ratio = (side ratio)² — don't forget to square the ratio
- Pythagoras theorem proof using similar triangles (not just formula) may be asked
- For proofs: state which similarity criterion you used and write the correspondence correctly (△ABC ~ △PQR means A↔P, B↔Q, C↔R)
- Always check if triplet is Pythagorean before concluding right angle
📚 RD Sharma Mathematics Class 10 — Triangles with Proofs
📝 Oswaal CBSE Maths Question Bank — Geometry Proofs