Congruence and Similarity of Triangles
Congruence and Similarity of Triangles
Congruent triangles are exact copies; similar triangles have the same shape at a different scale. For NTPC, the key results are the congruence rules and, above all, the scaling rules for similar triangles: sides, altitudes, medians and perimeters scale by k, and areas by k².
1. Criteria Box
| Congruence rule | What must match |
|---|---|
| SSS | Three sides |
| SAS | Two sides and the angle between them |
| ASA | Two angles and the side between them |
| AAS | Two angles and a side not between them |
| RHS | Right angle, hypotenuse and one side |
| Not valid | SSA (angle not included) and AAA (gives similarity only) |
| Similarity rule | What must hold |
|---|---|
| AA | Two pairs of angles equal |
| SSS | All three side ratios equal |
| SAS | Two side ratios equal and the included angles equal |
2. Scaling Rules for Similar Triangles (side ratio k)
| Quantity | Ratio |
|---|---|
| Sides, perimeters, altitudes, medians, angle bisectors, inradius, circumradius | k |
| Areas | k² |
| Angles | Equal (1 : 1) |
Basic proportionality (Thales): if DE ∥ BC in triangle ABC, then AD/DB = AE/EC and triangle ADE ∼ triangle ABC
Right triangle with altitude AD to the hypotenuse BC: AD² = BD × DC, AB² = BD × BC, AC² = DC × BC, AD = AB × AC/BC
✗ Areas of similar triangles are 196 and 324, so altitudes are in ratio 196 : 324 | ✓ Take square roots: 14 : 18 = 7 : 9
✗ AAA proves congruence | ✓ AAA proves only similarity; sizes can differ
हिंदी नोट: समरूप त्रिभुजों में भुजाओं, ऊँचाइयों, माध्यिकाओं और परिमाप का अनुपात k हो तो क्षेत्रफलों का अनुपात k² होता है। सर्वांगसमता के नियम: SSS, SAS, ASA, AAS और RHS; SSA और AAA मान्य नहीं।
Exam Pointer: Verified NTPC patterns: ratio of altitudes from the areas of similar triangles (March 2026 Graduate CBT-1) and ratio of areas from corresponding sides (June 2025 Graduate CBT-1). Congruence rules, Thales and right-triangle altitude questions had no verified NTPC shift and are tagged syllabus-based.
Pariksha Pattern: Every Way NTPC Asks This Topic
Pattern 1: Congruence rules and matching parts
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. In triangles ABC and PQR, AB = PQ, ∠B = ∠Q and BC = QR. Which rule makes them congruent, and which side equals AC?
Two sides and the included angle match, so SAS. Corresponding parts follow the letter order: AC = PR.
Answer: SAS; PR
EXAMATLAS LEVEL
Q. Which of SSS, SAS, ASA, AAS, SSA, AAA and RHS do not guarantee congruence? If triangle ABC ≅ triangle DEF with ∠A = 50° and ∠E = 60°, find ∠F.
SSA and AAA do not. Under ABC ↔ DEF, ∠B = ∠E = 60° and ∠D = ∠A = 50°, so ∠F = 180° − 50° − 60° = 70°.
Answer: SSA and AAA; 70°
Pattern 2: Similar triangles: areas, sides and altitudes
[PYQ: NTPC Graduate CBT-1 22-Mar-2026 Shift-1 | NTPC Graduate CBT-1 9-Jun-2025 Shift-1]
EXAM LEVEL
Q. Corresponding sides of two similar triangles are in the ratio 3 : 5. Find the ratio of their areas and of their perimeters.
Areas 9 : 25 and perimeters 3 : 5.
Answer: 9 : 25 and 3 : 5
EXAMATLAS LEVEL
Q. The areas of two similar triangles are 196 cm² and 324 cm². An altitude of the larger is 27 cm. Find the corresponding altitude of the smaller, and the ratio of their medians.
Side ratio = √196 : √324 = 14 : 18 = 7 : 9. Altitude of the smaller = 27 × 7/9 = 21 cm. Medians are also in the ratio 7 : 9.
Answer: 21 cm; 7 : 9
Pattern 3: Basic proportionality (a line parallel to one side)
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. In triangle ABC, DE ∥ BC with D on AB and E on AC. AD = 3 cm, DB = 5 cm and AE = 4.5 cm. Find EC.
AD/DB = AE/EC gives 3/5 = 4.5/EC, so EC = 7.5 cm.
Answer: 7.5 cm
EXAMATLAS LEVEL
Q. In triangle ABC, DE ∥ BC with AD : DB = 2 : 3. The area of trapezium DECB is 84 cm². Find the area of triangle ADE.
AD : AB = 2 : 5, so area ADE : area ABC = 4 : 25 and the trapezium is 21 parts. 21 parts = 84, so 1 part = 4 and area ADE = 16 cm².
Answer: 16 cm²
Pattern 4: Altitude to the hypotenuse of a right triangle
[Pattern: syllabus-based, PYQ-style]
EXAM LEVEL
Q. Triangle ABC is right-angled at A and AD ⊥ BC. BD = 4 cm and DC = 9 cm. Find AD.
AD² = BD × DC = 36, so AD = 6 cm.
Answer: 6 cm
EXAMATLAS LEVEL
Q. Triangle ABC is right-angled at A with AB = 15 cm and AC = 20 cm, and AD ⊥ BC. Find AD, BD and DC.
BC = 25 cm. AD = 15 × 20/25 = 12 cm. BD = AB²/BC = 225/25 = 9 cm and DC = 400/25 = 16 cm. Check: 9 × 16 = 144 = 12².
Answer: 12 cm, 9 cm, 16 cm
60-Second Revision
- Congruence: SSS, SAS, ASA, AAS, RHS; never SSA or AAA.
- Similar triangles: every length scales by k, area by k².
- DE ∥ BC: AD/DB = AE/EC; area of ADE = (AD/AB)² × area of ABC.
- Right triangle altitude: AD² = BD × DC.