Criteria for Similarity of Triangles
The criteria for similarity of triangles form a central topic in Chapter 6 (Triangles) of the CBSE Class 10 Mathematics syllabus. While the definition of similar triangles requires that all three pairs of corresponding angles be equal AND all three pairs of corresponding sides be in the same ratio, checking all six conditions every time would be cumbersome. The similarity criteria provide shortcuts: they tell us that checking just two or three specific conditions is sufficient to establish that two triangles are similar. There are three main criteria for establishing similarity between triangles: the AA (Angle-Angle) Criterion, the SSS (Side-Side-Side) Similarity Criterion, and the SAS (Side-Angle-Side) Similarity Criterion. Each criterion reduces the amount of information needed to conclude similarity, making proofs and computations far more efficient. These criteria are analogous to the congruence criteria (SAS, ASA, SSS, RHS) but differ in that they deal with proportional sides rather than equal sides. The similarity criteria are built upon the Basic Proportionality Theorem (Thales' Theorem) and are used extensively in geometry problems involving indirect measurement, shadow calculations, map-making, and architectural design. Mastering these criteria is essential not only for board examinations but also for building a strong foundation in geometric reasoning that extends to trigonometry, coordinate geometry, and higher mathematics. This guide provides detailed explanations, proofs, comparison tables, and worked examples for each criterion.
What is Criteria for Similarity of Triangles - AA, SSS, SAS, Proofs & Examples?
Similar Triangles: Two triangles are similar if (i) their corresponding angles are equal, and (ii) their corresponding sides are in the same ratio. The notation triangle ABC ~ triangle DEF means A corresponds to D, B corresponds to E, C corresponds to F.
The Three Similarity Criteria:
1. AA (Angle-Angle) Criterion: If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
If Angle A = Angle D and Angle B = Angle E, then triangle ABC ~ triangle DEF.
Note: Since the sum of angles of a triangle is 180 degrees, if two pairs of angles are equal, the third pair must also be equal. Therefore, the AA criterion is sometimes stated as AAA (Angle-Angle-Angle), but checking two angles is sufficient.
2. SSS (Side-Side-Side) Similarity Criterion: If in two triangles, the corresponding sides are in the same ratio (i.e., all three pairs of sides are proportional), then the two triangles are similar.
If AB/DE = BC/EF = CA/FD, then triangle ABC ~ triangle DEF.
3. SAS (Side-Angle-Side) Similarity Criterion: If one angle of a triangle is equal to one angle of the other triangle, and the sides including these angles are proportional, then the two triangles are similar.
If Angle A = Angle D and AB/DE = AC/DF, then triangle ABC ~ triangle DEF.
Note: The equal angle must be the included angle between the two proportional sides. If the equal angle is not the included angle, the criterion does not apply.
Comparison with Congruence Criteria:
| Aspect | Congruence | Similarity |
|---|---|---|
| Definition | Same shape AND same size | Same shape, possibly different size |
| Sides condition | Corresponding sides are equal | Corresponding sides are proportional |
| Criteria | SAS, ASA, SSS, RHS | AA, SSS, SAS |
| Scale factor | Always 1 | Any positive number |
| Notation | Triangle ABC congruent to triangle DEF | Triangle ABC ~ triangle DEF |
Criteria for Similarity of Triangles Formula
Similarity Criteria Summary:
| Criterion | Conditions Required | Conclusion |
|---|---|---|
| AA Criterion | Two pairs of corresponding angles are equal: Angle A = Angle D, Angle B = Angle E | Triangle ABC ~ Triangle DEF |
| SSS Criterion | All three pairs of corresponding sides are proportional: AB/DE = BC/EF = CA/FD | Triangle ABC ~ Triangle DEF |
| SAS Criterion | Two pairs of sides proportional and included angles equal: AB/DE = AC/DF, Angle A = Angle D | Triangle ABC ~ Triangle DEF |
Derived Results from Similarity:
If triangle ABC ~ triangle DEF with scale factor k = AB/DE, then:
| Property | Ratio |
|---|---|
| Corresponding sides | AB/DE = BC/EF = CA/FD = k |
| Corresponding angles | Angle A = Angle D, Angle B = Angle E, Angle C = Angle F |
| Perimeters | Perimeter(ABC)/Perimeter(DEF) = k |
| Altitudes | Altitude from A / Altitude from D = k |
| Medians | Median from A / Median from D = k |
| Areas | Area(ABC)/Area(DEF) = k^2 |
Derivation and Proof
Proof of the AA Similarity Criterion:
Given: Two triangles ABC and DEF such that Angle A = Angle D and Angle B = Angle E.
To Prove: Triangle ABC ~ Triangle DEF, that is, AB/DE = BC/EF = CA/FD.
Construction: On ray DE, mark a point P such that DP = AB. On ray DF, mark a point Q such that DQ = AC. Join PQ.
Step 1: In triangles ABC and DPQ: AB = DP (construction), Angle A = Angle D (given), AC = DQ (construction). By SAS congruence, triangle ABC is congruent to triangle DPQ.
Step 2: From the congruence: Angle B = Angle DPQ. But Angle B = Angle E (given). Therefore Angle DPQ = Angle E.
Step 3: In triangle DEF, since Angle DPQ = Angle DEF, and these are corresponding angles with PQ as a transversal, PQ is parallel to EF.
Step 4: By BPT applied to triangle DEF with PQ || EF: DP/DE = DQ/DF.
Step 5: Since DP = AB and DQ = AC: AB/DE = AC/DF. ... (i)
Step 6: Similarly, by taking construction on other sides, we get AB/DE = BC/EF. ... (ii)
Step 7: From (i) and (ii): AB/DE = BC/EF = CA/FD. Combined with equal angles, triangle ABC ~ triangle DEF. QED.
Proof of SSS Similarity Criterion:
Given: In triangles ABC and DEF, AB/DE = BC/EF = CA/FD.
To Prove: Triangle ABC ~ Triangle DEF.
Construction: On ray DE, mark P such that DP = AB. On ray DF, mark Q such that DQ = AC. Join PQ.
Step 1: AB/DE = AC/DF (given). Since DP = AB and DQ = AC: DP/DE = DQ/DF.
Step 2: By the converse of BPT, PQ || EF in triangle DEF.
Step 3: Since PQ || EF, triangle DPQ ~ triangle DEF (by AA: angle D common, angle DPQ = angle DEF).
Step 4: So DP/DE = PQ/EF. Since DP = AB: AB/DE = PQ/EF.
Step 5: But AB/DE = BC/EF (given). So PQ/EF = BC/EF, giving PQ = BC.
Step 6: In triangles ABC and DPQ: AB = DP, AC = DQ, BC = PQ. By SSS congruence, triangle ABC is congruent to triangle DPQ.
Step 7: Since triangle DPQ ~ triangle DEF, and triangle ABC is congruent to triangle DPQ, therefore triangle ABC ~ triangle DEF. QED.
Proof of SAS Similarity Criterion:
Given: In triangles ABC and DEF, AB/DE = AC/DF and Angle A = Angle D.
To Prove: Triangle ABC ~ Triangle DEF.
Construction: Same as above, mark P on DE with DP = AB and Q on DF with DQ = AC.
Step 1: In triangles ABC and DPQ: AB = DP, Angle A = Angle D, AC = DQ. By SAS congruence, triangle ABC is congruent to triangle DPQ.
Step 2: Since DP/DE = DQ/DF (from given ratio and construction), by converse of BPT, PQ || EF.
Step 3: Therefore triangle DPQ ~ triangle DEF (by AA). Since triangle ABC is congruent to triangle DPQ, we get triangle ABC ~ triangle DEF. QED.
Types and Properties
The three similarity criteria lead to different types of problems and geometric configurations:
1. AA Criterion Problems:
These are the most common. You identify two pairs of equal angles using properties like vertically opposite angles, alternate interior angles (with parallel lines), corresponding angles, common angles, or angles in the same segment of a circle.
Typical configurations: parallel lines cutting two sides of a triangle, X-shaped figures with vertically opposite angles, right triangles sharing an acute angle, tangent-radius configurations.
2. SSS Similarity Problems:
You are given all six side lengths of two triangles and must check if the three ratios are equal. Arrange the sides of each triangle in ascending (or descending) order, pair them up, and check ratios.
Typical configurations: two triangles with all measurements given, problems asking to identify similar triangles from a set of given side lengths.
3. SAS Similarity Problems:
You are given two sides and the included angle for each triangle. Check if the two sides are proportional and the included angles are equal.
Typical configurations: two triangles sharing a common angle (like an angle of a parallelogram), isosceles triangles with the vertex angle given, configurations where one angle is clearly identified as the included angle.
Common Geometric Configurations for Similarity:
| Configuration | Criterion Used | Key Feature |
|---|---|---|
| Parallel line within a triangle | AA | Corresponding/alternate angles from parallel lines |
| Two secants from an external point | AA | Common angle + vertically opposite angles |
| Altitude from right angle to hypotenuse | AA | Common angle + right angle |
| Tangent-secant from external point | AA | Tangent-chord angle = inscribed angle |
| Two triangles with given side lengths | SSS | Check all three ratios |
| Triangles sharing a vertex angle | SAS | Common angle + proportional adjacent sides |
Methods
Method 1: Establishing Similarity Using AA
Step 1: Identify two pairs of equal angles using geometric properties (parallel lines, vertically opposite angles, common angles, right angles). Step 2: State the AA criterion. Step 3: Write the similarity statement with correct vertex correspondence.
Example: In triangle ABC, DE || BC with D on AB and E on AC. Angle ADE = Angle ABC (corresponding angles). Angle A is common. By AA, triangle ADE ~ triangle ABC.
Method 2: Establishing Similarity Using SSS
Step 1: List all six sides. Step 2: Arrange each triangle's sides in order. Step 3: Compute the three ratios of corresponding sides. Step 4: If all ratios are equal, state SSS similarity with correct correspondence.
Example: Triangle ABC has sides 3, 4, 5. Triangle DEF has sides 6, 8, 10. Ratios: 3/6 = 4/8 = 5/10 = 1/2. By SSS, triangle ABC ~ triangle DEF.
Method 3: Establishing Similarity Using SAS
Step 1: Identify one pair of equal angles. Step 2: Check that the sides including this angle are proportional. Step 3: State SAS similarity.
Example: In triangles ABC and DEF, Angle B = Angle E = 70 degrees. AB/DE = 6/9 = 2/3, BC/EF = 8/12 = 2/3. Since AB/DE = BC/EF and the included angle B = angle E, by SAS, triangle ABC ~ triangle DEF.
Method 4: Finding Unknown Sides Using Similarity
Once similarity is established, use the equal ratios of corresponding sides to find unknowns. Set up the proportion and solve.
Method 5: Finding Unknown Angles Using Similarity
Once similarity is established, equate corresponding angles. Use the angle sum property (180 degrees) if needed.
Important Tips:
- Always write the similarity statement with the correct order of vertices. A corresponds to D, B to E, C to F in triangle ABC ~ triangle DEF.
- For SSS, arrange sides in ascending order in both triangles before computing ratios to ensure correct pairing.
- For SAS, verify that the equal angle is the INCLUDED angle between the two proportional sides. A non-included equal angle does not guarantee similarity.
- The AA criterion is the most commonly used because angles are often easier to identify than side lengths in geometric figures.
Solved Examples
Example 1: Identifying Similarity Using AA Criterion
Problem: In the figure, PQ || RS. Angle QPR = 50 degrees and Angle PRS = 30 degrees. Show that triangle PQR ~ triangle SRP.
Solution:
Step 1: Since PQ || RS, and PR is a transversal: Angle QPR = Angle PRS (alternate interior angles). Wait, we are given angle QPR = 50 and angle PRS = 30, which are not equal. Let us re-examine.
Step 2: Let us reconsider the figure. In triangles PQR and SRP, with PQ || RS and QR as a transversal: Angle PQR = Angle QRS (alternate interior angles, since PQ || RS). Also, Angle QPR = Angle PSR (alternate interior angles, since PQ || RS with PR as transversal).
Step 3: Thus we have two pairs of equal angles. By the AA criterion, triangle PQR ~ triangle SRP.
Answer: Triangle PQR ~ triangle SRP (by AA similarity).
Example 2: Checking SSS Similarity
Problem: Triangle ABC has sides AB = 5 cm, BC = 6 cm, and CA = 7 cm. Triangle PQR has sides PQ = 10 cm, QR = 12 cm, and RP = 14 cm. Check whether the triangles are similar.
Solution:
Step 1: Arrange the sides of each triangle: ABC: 5, 6, 7 and PQR: 10, 12, 14.
Step 2: Compute ratios: AB/PQ = 5/10 = 1/2, BC/QR = 6/12 = 1/2, CA/RP = 7/14 = 1/2.
Step 3: Since AB/PQ = BC/QR = CA/RP = 1/2, all three ratios are equal.
Step 4: By the SSS similarity criterion, triangle ABC ~ triangle PQR with scale factor 1/2.
Answer: Yes, triangle ABC ~ triangle PQR by SSS similarity with scale factor 1:2.
Example 3: Verifying SAS Similarity
Problem: In triangle LMN, LM = 6 cm, LN = 9 cm, and Angle L = 80 degrees. In triangle XYZ, XY = 4 cm, XZ = 6 cm, and Angle X = 80 degrees. Are the triangles similar?
Solution:
Step 1: The included angle (between the given sides): Angle L = Angle X = 80 degrees. The given sides include this angle in both triangles.
Step 2: Check proportionality: LM/XY = 6/4 = 3/2 and LN/XZ = 9/6 = 3/2.
Step 3: Since LM/XY = LN/XZ = 3/2 and the included angle L = angle X, by SAS similarity criterion, triangle LMN ~ triangle XYZ.
Answer: Yes, triangle LMN ~ triangle XYZ by SAS similarity with scale factor 3:2.
Example 4: Using AA to Prove Similarity in an Altitude Configuration
Problem: In triangle ABC, angle A = 90 degrees. AD is the altitude from A to BC. Prove that (i) triangle ABD ~ triangle CBA, and (ii) triangle ACD ~ triangle BCA.
Solution:
Part (i): In triangles ABD and CBA:
Angle ADB = 90 degrees = Angle BAC (both are right angles).
Angle ABD = Angle CBA (common angle B).
By AA criterion, triangle ABD ~ triangle CBA.
Part (ii): In triangles ACD and BCA:
Angle ADC = 90 degrees = Angle BAC (both are right angles).
Angle ACD = Angle BCA (common angle C).
By AA criterion, triangle ACD ~ triangle BCA.
Conclusion: Both parts are proved. Note: This also implies triangle ABD ~ triangle ACD, and from these similarities we can derive the Pythagoras Theorem.
Example 5: Finding Unknown Sides Using Similarity
Problem: Triangle ABC ~ Triangle DEF. AB = 4 cm, BC = 5 cm, CA = 6 cm, and DE = 6 cm. Find EF and FD.
Solution:
Step 1: Since triangle ABC ~ triangle DEF, the ratios of corresponding sides are equal: AB/DE = BC/EF = CA/FD.
Step 2: AB/DE = 4/6 = 2/3.
Step 3: BC/EF = 2/3, so EF = 5 x 3/2 = 7.5 cm.
Step 4: CA/FD = 2/3, so FD = 6 x 3/2 = 9 cm.
Answer: EF = 7.5 cm and FD = 9 cm.
Example 6: Similarity in a Trapezium
Problem: In trapezium ABCD, AB || DC. Diagonals intersect at O. If AB = 12 cm and DC = 8 cm, find the ratio OA:OC.
Solution:
Step 1: In triangles AOB and COD: Angle AOB = Angle COD (vertically opposite angles). Angle OAB = Angle OCD (alternate interior angles, since AB || DC).
Step 2: By AA criterion, triangle AOB ~ triangle COD.
Step 3: OA/OC = AB/CD = 12/8 = 3/2.
Answer: OA:OC = 3:2.
Example 7: When SSS Similarity Fails
Problem: Triangle PQR has sides 3, 5, 7. Triangle STU has sides 6, 10, 15. Are they similar?
Solution:
Step 1: Arrange in ascending order: PQR: 3, 5, 7 and STU: 6, 10, 15.
Step 2: Compute ratios: 3/6 = 1/2, 5/10 = 1/2, 7/15 = 7/15.
Step 3: 1/2 is not equal to 7/15 (since 1/2 = 15/30 and 7/15 = 14/30).
Step 4: Since not all ratios are equal, the triangles are NOT similar by SSS.
Answer: No, the triangles are not similar.
Example 8: SAS Similarity with a Common Angle
Problem: In the figure, two triangles share a common vertex at A. Triangle ABE has AB = 10 cm, AE = 8 cm, and triangle ACD has AC = 15 cm, AD = 12 cm. If angle BAE = angle CAD, show that triangle ABE ~ triangle ACD.
Solution:
Step 1: Check proportionality of sides including the equal angle: AB/AC = 10/15 = 2/3 and AE/AD = 8/12 = 2/3.
Step 2: AB/AC = AE/AD = 2/3, and the included angle BAE = angle CAD (given).
Step 3: By SAS similarity criterion, triangle ABE ~ triangle ACD.
Answer: Triangle ABE ~ triangle ACD (by SAS similarity).
Example 9: Altitude and Similar Triangles
Problem: In triangle PQR, angle Q = 90 degrees. QS is perpendicular to PR. If PS = 6 cm and SR = 8 cm, find QS, PQ, and QR.
Solution:
Step 1: PR = PS + SR = 6 + 8 = 14 cm.
Step 2: Triangle PQS ~ triangle PQR (AA: angle P is common, angle PSQ = angle PQR = 90 degrees). Wait, angle PQR = 90 degrees, and angle PSQ = 90 degrees, so this works differently. Let us use the standard results.
Step 3: In the altitude-on-hypotenuse configuration: QS^2 = PS x SR = 6 x 8 = 48. So QS = sqrt(48) = 4sqrt(3) cm.
Step 4: PQ^2 = PS x PR = 6 x 14 = 84. So PQ = sqrt(84) = 2sqrt(21) cm.
Step 5: QR^2 = SR x PR = 8 x 14 = 112. So QR = sqrt(112) = 4sqrt(7) cm.
Verification: PQ^2 + QR^2 = 84 + 112 = 196 = 14^2 = PR^2. Correct by Pythagoras Theorem.
Answer: QS = 4sqrt(3) cm, PQ = 2sqrt(21) cm, QR = 4sqrt(7) cm.
Example 10: Proving a Geometric Result Using Similarity
Problem: In triangle ABC, the bisector of angle A meets BC at D. Prove that BD/DC = AB/AC.
Solution:
Step 1: Draw a line CE parallel to DA, meeting BA extended at E.
Step 2: Since CE || DA and AC is a transversal: Angle DAC = Angle ACE (alternate interior angles). ... (i)
Step 3: Since CE || DA and BE is a transversal: Angle BAD = Angle AEC (corresponding angles). ... (ii)
Step 4: Since AD bisects angle A: Angle BAD = Angle DAC. ... (iii)
Step 5: From (i), (ii), and (iii): Angle ACE = Angle AEC. Therefore triangle ACE is isosceles with AC = AE. ... (iv)
Step 6: In triangle BCE, DA || CE (by construction). By BPT: BD/DC = BA/AE. ... (v)
Step 7: From (iv) and (v): BD/DC = BA/AC = AB/AC.
Conclusion: BD/DC = AB/AC. This is the Angle Bisector Theorem, proved using the BPT and properties of parallel lines. QED.
Real-World Applications
The criteria for similarity of triangles have extensive applications in mathematics and the real world.
Indirect Height Measurement: One of the oldest practical applications of similar triangles is measuring the height of tall objects like buildings, towers, trees, and mountains. By observing the shadow cast by a stick of known height and comparing it to the shadow of the tall object (both measured at the same time so that the sun's angle is identical), similar triangles are formed. The AA criterion applies because the sun's rays are parallel, creating equal angles of elevation. The unknown height is then found by proportion.
River Width Measurement: Surveyors standing on one bank of a river can measure its width without crossing it. By setting up a triangle with a known baseline on their bank and measuring angles to a point on the opposite bank, they create a triangle similar to a smaller reference triangle. The width is then calculated using proportionality.
Scale Models and Maps: Every architectural model and every map is a practical application of similar triangles. The scale factor (like 1:100 for a model or 1:50000 for a map) means that every triangle on the model or map is similar to the corresponding triangle in reality. The SSS similarity criterion guarantees that if all distances are scaled by the same factor, the shapes are preserved.
Photography and Lens Optics: When a camera lens forms an image, the object and the image form similar triangles. The magnification ratio equals the ratio of image distance to object distance, which is derived from similar triangle relationships. This is fundamental in optics, telescope design, and microscopy.
Engineering and Machine Design: Gear ratios, pulley systems, and lever mechanisms often involve proportional relationships derivable from similar triangles. The SAS criterion is particularly useful in structural engineering where angles and adjacent member lengths are the primary design parameters.
Computer Vision and Image Recognition: Algorithms for detecting objects in images use the properties of similar triangles. When an object moves closer to or further from a camera, its image on the sensor changes size proportionally. By recognising this similarity transformation, software can estimate distance and object size, enabling applications from autonomous vehicles to facial recognition.
Astronomy: The distance to nearby stars is measured using parallax, which involves similar triangles. The baseline is the diameter of Earth's orbit, and the tiny angle shift of the star creates a very elongated similar triangle whose proportions reveal the stellar distance.
Key Points to Remember
- Two triangles are similar if their corresponding angles are equal and their corresponding sides are in the same ratio.
- There are three criteria for similarity: AA, SSS, and SAS.
- AA (Angle-Angle): If two pairs of corresponding angles are equal, the triangles are similar. This is the most commonly used criterion.
- SSS (Side-Side-Side): If all three pairs of corresponding sides are proportional, the triangles are similar.
- SAS (Side-Angle-Side): If two pairs of sides are proportional and the included angle is equal, the triangles are similar. The angle MUST be the included angle.
- The order of vertices in the similarity statement matters: triangle ABC ~ triangle DEF means A corresponds to D, B to E, C to F.
- All three criteria are proved using the Basic Proportionality Theorem (BPT) and its converse.
- If two triangles are similar with scale factor k, the ratio of their areas is k^2.
- The ratio of corresponding altitudes, medians, and perimeters of similar triangles equals the scale factor k.
- Similarity is different from congruence: congruence requires equal sides, similarity only requires proportional sides.
- The altitude from the right angle to the hypotenuse of a right triangle creates two triangles, each similar to the original and to each other.
- For CBSE board exams, be prepared to prove the AA criterion, apply all three criteria in numerical problems, and use similarity to find unknown sides and areas.
Practice Problems
- In triangle ABC, angle A = 65 degrees, angle B = 75 degrees. In triangle DEF, angle D = 65 degrees, angle F = 40 degrees. Are the triangles similar? If yes, write the correct similarity statement.
- Triangle PQR has sides PQ = 8, QR = 10, PR = 12. Triangle STU has sides ST = 12, TU = 15, SU = 18. Check for SSS similarity.
- In triangles ABC and DEF, AB = 6, AC = 10, angle A = 50 degrees, DE = 3, DF = 5, angle D = 50 degrees. Prove similarity and find the scale factor.
- In right triangle ABC (angle B = 90 degrees), BD is the altitude from B to AC. If AD = 5 and DC = 12, find BD, AB, and BC.
- Two poles of height 6 m and 11 m stand 12 m apart on level ground. Find the distance between their tops and the point on the ground between them where the lines from their tops to the base of the other pole intersect.
- In triangle ABC, DE || BC with D on AB and E on AC. If AD/AB = 3/5, find (i) DE/BC, (ii) Area(ADE)/Area(ABC), (iii) Area(BCED)/Area(ABC).
- Prove the SAS similarity criterion using the BPT.
- State and prove the AA similarity criterion.
Frequently Asked Questions
Q1. What are the criteria for similarity of triangles?
There are three criteria: (1) AA (Angle-Angle) - if two pairs of corresponding angles are equal, (2) SSS (Side-Side-Side) - if all three pairs of corresponding sides are proportional, (3) SAS (Side-Angle-Side) - if two pairs of sides are proportional and the included angle is equal. Any one criterion is sufficient to establish similarity.
Q2. Why is there no ASA similarity criterion?
The ASA condition for similarity would mean two angles and the included side are involved. However, two pairs of equal angles (AA) alone are already sufficient for similarity, making the side information redundant. So ASA is not needed as a separate criterion because it is subsumed by the AA criterion.
Q3. What is the difference between similarity and congruence criteria?
For congruence, corresponding sides must be equal; for similarity, they must be proportional. Congruence criteria (SAS, ASA, SSS, RHS) require equal sides. Similarity criteria (AA, SSS, SAS) require proportional sides. Congruence implies similarity with scale factor 1, but similarity does not imply congruence.
Q4. Why must the angle be the included angle in SAS similarity?
If the equal angle is not between the two proportional sides, the triangles may not be similar. Consider two triangles where a non-included angle is equal and two sides are proportional: different shapes can result because the angle does not constrain the triangle uniquely. Only the included angle locks the shape.
Q5. How do you determine the correct vertex correspondence?
Match each angle of one triangle with the equal angle of the other. The vertex of each angle defines the correspondence. For example, if angle A = angle D, angle B = angle E, angle C = angle F, write triangle ABC ~ triangle DEF. The order of vertices must reflect the angle matching.
Q6. Can two equilateral triangles of different sizes be similar?
Yes. All equilateral triangles have three 60-degree angles. By the AA criterion, any two equilateral triangles are similar. The scale factor is the ratio of their side lengths. This is a good example showing that similarity preserves shape but allows different sizes.
Q7. How are similarity criteria used to prove the Pythagoras Theorem?
In a right triangle ABC with angle B = 90 degrees, the altitude BD to hypotenuse AC creates three similar triangles: triangle ABD ~ triangle BCD ~ triangle ABC (by AA criterion in each case). Using the proportional sides from these similarities gives AB^2 = AD.AC and BC^2 = CD.AC. Adding: AB^2 + BC^2 = AC(AD + CD) = AC^2, which is the Pythagoras Theorem.
Q8. What is the most important similarity criterion for CBSE Class 10 exams?
The AA (Angle-Angle) criterion is the most frequently used and tested in CBSE exams. Most geometry problems in the Triangles chapter require identifying two pairs of equal angles to establish similarity. The proof of the AA criterion is also commonly asked as a board exam question worth 3-5 marks.
Related Topics
- AA Similarity Criterion
- SSS Similarity Criterion
- SAS Similarity Criterion
- Similar Triangles
- Angle Sum Property of Triangle
- Exterior Angle Property of Triangle
- Properties of Isosceles Triangle
- Properties of Equilateral Triangle
- Triangle Inequality Property
- Medians and Altitudes of Triangle
- Right-Angled Triangle Property
- Congruent Triangles - Proofs
- Inequalities in Triangles
- Basic Proportionality Theorem (BPT)










