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SAS Similarity Criterion

Class 10Triangles

The SAS (Side-Angle-Side) Similarity Criterion is one of the three criteria for proving similarity of triangles, covered in Chapter 6 (Triangles) of the NCERT Class 10 Mathematics textbook.


This criterion states that:

  • If one angle of a triangle is equal to one angle of another triangle, and the sides including these angles are proportional, then the two triangles are similar.

The SAS similarity criterion combines both angle equality and side proportionality. It is particularly useful when you know one pair of angles is equal and need to check only two pairs of sides (not all three, as in SSS).



The SAS criterion is particularly powerful because it requires checking only one angle and two sides, making it efficient for both numerical problems and geometric proofs. In many practical problems — such as determining whether scale models are accurate representations of buildings, or whether shadows cast by similar objects maintain proportional relationships — the SAS criterion provides the quickest route to establishing similarity.



In the NCERT framework, the SAS similarity criterion is presented as Theorem 6.5, and its proof relies on the converse of the Basic Proportionality Theorem (BPT). Students are expected to understand both the statement and the proof of this criterion, as it frequently appears in CBSE board examinations.

What is SAS Similarity Criterion — Statement, Proof & Solved Examples?

SAS Similarity Criterion:

If one angle of a triangle is equal to one angle of another triangle and the sides including these equal angles are proportional, then the two triangles are similar.


Formal statement: If in triangles ABC and DEF:

  • Angle A = Angle D
  • AB/DE = AC/DF (sides including the equal angles are proportional)

then triangle ABC is similar to triangle DEF.


Key points:

  • The equal angle must be the angle included between the two proportional sides.
  • If the angle is NOT between the proportional sides, SAS similarity does NOT apply.
  • SAS similarity is different from SAS congruence — congruence requires sides to be equal, similarity requires proportionality.

Comparison of the three similarity criteria:

CriterionWhat You Need
AATwo pairs of equal angles
SSSAll three pairs of sides proportional
SASOne pair of equal angles + the two including sides proportional

Comparison of all three similarity criteria:

CriterionConditions RequiredAbbreviation
AA SimilarityTwo pairs of corresponding angles are equalAA or AAA
SSS SimilarityAll three pairs of corresponding sides are proportionalSSS
SAS SimilarityOne pair of equal angles with the including sides proportionalSAS

Important distinction: In SAS congruence, the sides must be equal. In SAS similarity, the sides must be proportional. This is a common source of confusion that must be avoided in examinations.

SAS Similarity Criterion Formula

SAS Similarity — Conditions:

Angle A = Angle D
AND
AB/DE = AC/DF

implies triangle ABC is similar to triangle DEF


Once similarity is established:

  • All corresponding angles are equal: angle A = angle D, angle B = angle E, angle C = angle F.
  • All corresponding sides are proportional: AB/DE = BC/EF = AC/DF = k.
  • Ratio of perimeters = k.
  • Ratio of areas = k².

Important: The included angle is critical. Check that the angle lies between the two sides being compared.

Derivation and Proof

Proof of SAS Similarity Criterion:


Given: In triangles ABC and DEF, angle A = angle D and AB/DE = AC/DF.

To prove: Triangle ABC is similar to triangle DEF.


Construction: On DE, mark point P such that DP = AB. On DF, mark point Q such that DQ = AC. Join PQ.


Proof:

  1. In triangles ABC and DPQ:
    • AB = DP (by construction)
    • AC = DQ (by construction)
    • Angle A = Angle D (given)
  2. By SAS congruence: triangle ABC is congruent to triangle DPQ.
  3. Therefore: angle B = angle DPQ and angle C = angle DQP.
  4. Now, from the given: AB/DE = AC/DF, and AB = DP, AC = DQ.
  5. So: DP/DE = DQ/DF.
  6. By the converse of BPT: PQ is parallel to EF.
  7. Since PQ is parallel to EF: angle DPQ = angle E and angle DQP = angle F (corresponding angles).
  8. From step 3: angle B = angle DPQ = angle E, and angle C = angle DQP = angle F.
  9. Therefore: angle A = angle D, angle B = angle E, angle C = angle F.
  10. All corresponding angles are equal, so triangle ABC is similar to triangle DEF.

QED.

Types and Properties

Types of SAS similarity problems:


Type 1: Verifying SAS similarity

  • Given one pair of equal angles and two pairs of sides, check whether the angle is included and the sides are proportional.

Type 2: Finding unknown sides

  • Given SAS similarity, find an unknown side using proportionality.

Type 3: Proving triangles similar

  • In a geometric figure, identify the equal angle and show that the including sides are proportional.

Type 4: Application with other results


Type 5: Area and perimeter ratios

  • Using SAS similarity to establish the scale factor, then computing area/perimeter ratios.

Methods

Step-by-step method:

  1. Identify the equal angle in both triangles.
  2. Identify the sides that include (form) this angle in each triangle.
  3. Compute the ratios of these including sides.
  4. If the ratios are equal, SAS similarity applies.
  5. Write the similarity statement with correct vertex correspondence.

Identifying the included angle:

  • If the equal angle is A in triangle ABC and D in triangle DEF, then the including sides are AB, AC in the first and DE, DF in the second.
  • The ratio to check is AB/DE and AC/DF (NOT BC/EF).

Common mistakes:

  • Using a non-included angle — the angle MUST be between the two proportional sides.
  • Confusing SAS similarity with SAS congruence.
  • Not checking that the angle is actually equal (just assumed).
  • Wrong correspondence of vertices.


Method 4: Finding Ratios of Areas, Perimeters, and Altitudes

Once two triangles are established as similar with scale factor k:

  • Ratio of corresponding sides = k
  • Ratio of corresponding altitudes = k
  • Ratio of corresponding medians = k
  • Ratio of corresponding angle bisectors = k
  • Ratio of perimeters = k
  • Ratio of areas = k^2
  • Ratio of circumradii = k

Method 5: Identifying the Included Angle in Complex Figures

In complex geometric figures, identify the included angle by tracing the two sides from a common vertex. The angle at that vertex, formed between the two proportional sides, is the included angle. Common situations:

  • In overlapping triangles: the common angle at the shared vertex is the included angle.
  • In intersecting lines: vertically opposite angles serve as equal included angles.
  • In cyclic quadrilaterals: angles subtended by the same arc may provide the equal included angles.

Solved Examples

Example 1: Basic SAS Similarity Check

Problem: In triangle ABC, angle A = 50°, AB = 6 cm, AC = 9 cm. In triangle DEF, angle D = 50°, DE = 4 cm, DF = 6 cm. Are the triangles similar?


Solution:

Given:

  • Angle A = angle D = 50°
  • AB = 6, AC = 9, DE = 4, DF = 6

Steps:

  1. Angle A = angle D = 50° (equal angles) ✓
  2. AB/DE = 6/4 = 3/2
  3. AC/DF = 9/6 = 3/2
  4. AB/DE = AC/DF = 3/2 (including sides proportional) ✓

By SAS similarity, triangle ABC is similar to triangle DEF.

Answer: Yes, they are similar (scale factor = 3/2).

Example 2: SAS Similarity — Not Similar

Problem: Triangle PQR: angle P = 60°, PQ = 8 cm, PR = 12 cm. Triangle XYZ: angle X = 60°, XY = 6 cm, XZ = 10 cm. Are they similar?


Solution:

Given:

  • Angle P = angle X = 60°
  • PQ = 8, PR = 12, XY = 6, XZ = 10

Steps:

  1. Angle P = angle X ✓
  2. PQ/XY = 8/6 = 4/3
  3. PR/XZ = 12/10 = 6/5
  4. 4/3 is not equal to 6/5

Answer: No, the triangles are NOT similar by SAS.

Example 3: Finding an Unknown Side

Problem: Triangle ABC is similar to triangle DEF by SAS (angle A = angle D). AB = 10, AC = 15, DE = 6. Find DF.


Solution:

Given:

  • AB/DE = AC/DF (SAS similarity)
  • AB = 10, AC = 15, DE = 6

Steps:

  1. Scale factor = AB/DE = 10/6 = 5/3
  2. AC/DF = 5/3 → DF = 15 × 3/5 = 9

Answer: DF = 9 cm.

Example 4: SAS Similarity in a Triangle with Parallel Line

Problem: In triangle ABC, DE is parallel to BC, where D is on AB and E is on AC. Prove that triangle ADE is similar to triangle ABC using SAS similarity.


Solution:

Given:

  • DE is parallel to BC in triangle ABC

Proof:

  1. Angle A is common to both triangles ADE and ABC.
  2. By BPT: AD/AB = AE/AC (sides including angle A are proportional).
  3. Angle A = angle A (common) and AD/AB = AE/AC.
  4. By SAS similarity: triangle ADE is similar to triangle ABC.

Hence proved.

Example 5: SAS Similarity — Vertically Opposite Angles

Problem: Two triangles share a common vertex. In triangles AOB and COD, angle O is vertically opposite. OA = 6, OB = 9, OC = 4, OD = 6. Prove similarity.


Solution:

Given:

  • Angle AOB = angle COD (vertically opposite angles)
  • OA = 6, OB = 9, OC = 4, OD = 6

Steps:

  1. Angle AOB = angle COD (vertically opposite) ✓
  2. OA/OC = 6/4 = 3/2
  3. OB/OD = 9/6 = 3/2
  4. OA/OC = OB/OD = 3/2 ✓

By SAS similarity, triangle AOB is similar to triangle COD.

Hence proved.

Example 6: Area Ratio Using SAS Similarity

Problem: Two triangles are similar by SAS criterion. The including sides of the equal angle are 8 cm, 12 cm in one triangle and 6 cm, 9 cm in the other. Find the ratio of their areas.


Solution:

Given:

  • Sides: 8, 12 and 6, 9

Steps:

  1. Scale factor: 8/6 = 4/3 (or 12/9 = 4/3)
  2. Ratio of areas = k² = (4/3)² = 16/9

Answer: Ratio of areas = 16:9.

Example 7: SAS Similarity with Angle Bisector

Problem: In triangle ABC, the bisector of angle A meets BC at D. If AB = 10 cm, AC = 15 cm, BD = 4 cm, and DC = 6 cm, show that triangles ABD and ACD share certain similarity properties.


Solution:

Given:

  • AD bisects angle A, so angle BAD = angle CAD.
  • AB = 10, AC = 15, BD = 4, DC = 6

Check:

  1. Angle BAD = angle CAD (AD is bisector)
  2. AB/AC = 10/15 = 2/3
  3. BD/DC = 4/6 = 2/3
  4. AB/AC = BD/DC ✓ (angle bisector theorem confirmed)

Note: Triangles ABD and ACD are NOT similar by SAS because the equal angle (angle A halved) is included between AB and AD (not AB and BD). However, the angle bisector theorem is verified: BD/DC = AB/AC = 2/3.

Answer: The angle bisector theorem is confirmed: BD/DC = AB/AC = 2/3.

Example 8: Identifying the Correct Included Angle

Problem: In triangle ABC, angle B = 70°, AB = 8, BC = 12. In triangle DEF, angle E = 70°, DE = 4, EF = 6. Check SAS similarity.


Solution:

Given:

  • Angle B = angle E = 70°
  • AB = 8, BC = 12 (sides including angle B)
  • DE = 4, EF = 6 (sides including angle E)

Steps:

  1. Angle B = angle E ✓ (the included angle)
  2. AB/DE = 8/4 = 2
  3. BC/EF = 12/6 = 2
  4. AB/DE = BC/EF = 2 ✓

By SAS similarity: triangle ABC is similar to triangle DEF (B maps to E).

Answer: Yes, the triangles are similar.

Example 9: SAS Similarity — Common Angle

Problem: In the figure, triangles PQS and PRT share angle P. PQ = 4, PR = 6, PS = 6, PT = 9. Prove triangle PQS is similar to triangle PRT.


Solution:

Given:

  • Angle P is common to both triangles.
  • PQ = 4, PR = 6, PS = 6, PT = 9

Steps:

  1. Angle P = angle P (common) ✓
  2. PQ/PR = 4/6 = 2/3
  3. PS/PT = 6/9 = 2/3
  4. PQ/PR = PS/PT = 2/3 ✓

By SAS similarity: triangle PQS is similar to triangle PRT.

Hence proved.

Example 10: Finding All Sides After SAS Similarity

Problem: Triangle ABC is similar to triangle DEF by SAS (angle C = angle F = 90°). AC = 8, BC = 6, DF = 12. Find DE, EF, and the hypotenuse AB.


Solution:

Given:

  • Angle C = angle F = 90°
  • AC = 8, BC = 6, DF = 12

Steps:

  1. Scale factor: AC/DF = 8/12 = 2/3
  2. BC/EF = 2/3 → EF = 6 × 3/2 = 9
  3. AB = √(AC² + BC²) = √(64 + 36) = √100 = 10
  4. AB/DE = 2/3 → DE = 10 × 3/2 = 15

Verification: DE = √(DF² + EF²) = √(144 + 81) = √225 = 15 ✓

Answer: DE = 15, EF = 9, AB = 10.

Real-World Applications

Applications of SAS Similarity:

  • Height and distance problems — finding heights of buildings or trees using shadow measurements (angle of elevation is common).
  • Mirror reflections — angle of incidence equals angle of reflection, creating similar triangles.
  • Navigation — determining distances using triangulation with a known angle.
  • Architecture — scaling designs while preserving angles.
  • Optics — lens and mirror problems where light rays form similar triangles.
  • Surveying — measuring inaccessible distances with known angles and proportional sides.


Detailed applications:

  • Shadow problems: At a given time, the sun's rays make equal angles with the ground. If a person of height h1 casts a shadow s1, and a building of height h2 casts a shadow s2, the triangles formed are similar by SAS (equal angle of elevation, proportional sides). This gives h1/h2 = s1/s2.
  • Photography: When a camera captures an image, the triangle formed by the lens, the object, and the image sensor is similar to the triangle formed by the lens, the sensor, and the image — by SAS similarity.
  • Model building: Architects build scale models maintaining the same angles but proportional lengths. SAS similarity ensures that any cross-section through the model corresponds proportionally to the actual structure.

Key Points to Remember

  • SAS Similarity: If one angle of a triangle equals one angle of another, and the sides including these angles are proportional, the triangles are similar.
  • The equal angle MUST be the included angle — the angle between the two proportional sides.
  • If the angle is not included, SAS similarity does NOT apply.
  • SAS similarity is proved using SAS congruence and the converse of BPT.
  • Do NOT confuse SAS similarity (proportional sides) with SAS congruence (equal sides).
  • Once SAS similarity is established, all corresponding angles are equal and all sides are proportional.
  • The scale factor k gives: ratio of perimeters = k, ratio of areas = k².
  • Common sources of the equal angle: common angle, vertically opposite angles, corresponding angles (parallel lines).
  • In CBSE exams, SAS similarity questions carry 2–4 marks. Show the equal angle and both side ratios clearly.
  • Always state the correspondence: "Triangle ABC is similar to triangle DEF by SAS" with matching vertices.

Practice Problems

  1. In triangle ABC, angle A = 40°, AB = 10, AC = 15. In triangle PQR, angle P = 40°, PQ = 6, PR = 9. Are the triangles similar?
  2. Triangle ABC is similar to triangle DEF by SAS. AB = 14, AC = 21, DE = 10. Find DF.
  3. In triangle PQR, S is on PQ and T is on PR such that angle P is common. PS = 3, PQ = 9, PT = 4, PR = 12. Prove that triangle PST is similar to triangle PQR.
  4. Two triangles have an angle of 90° each. The sides including the right angle are 5, 12 in one and 10, 24 in the other. Prove similarity and find area ratio.
  5. In quadrilateral ABCD, diagonals intersect at O. Angle AOB = angle COD. OA = 8, OB = 12, OC = 6, OD = 9. Prove that triangle AOB is similar to triangle COD.
  6. Can SAS similarity be applied if the equal angle is NOT between the proportional sides? Explain with an example.

Frequently Asked Questions

Q1. What is the SAS similarity criterion?

If one angle of a triangle equals one angle of another triangle, and the sides including (forming) these angles are proportional, then the triangles are similar.

Q2. Why must the angle be the INCLUDED angle?

The proof of SAS similarity relies on the converse of BPT, which requires the proportional sides to form the equal angle. If the angle is not between the proportional sides, the proportionality does not guarantee similarity.

Q3. How is SAS similarity different from SAS congruence?

SAS congruence requires two sides to be EQUAL and the included angle to be equal. SAS similarity requires two sides to be PROPORTIONAL and the included angle to be equal. Congruent triangles are identical; similar triangles have the same shape but may differ in size.

Q4. What are common sources of the equal angle in SAS similarity?

Common angle (two triangles sharing a vertex), vertically opposite angles (two triangles at an intersection), corresponding or alternate angles (parallel lines), and right angles.

Q5. Can I use SAS similarity if I know two angles are equal?

If two angles are equal, you should use AA similarity instead, which is simpler and does not require checking sides at all.

Q6. How many side ratios do I need to check for SAS?

Only TWO side ratios — the sides that include the equal angle. This is simpler than SSS which requires all three.

Q7. How do I find the scale factor from SAS similarity?

The scale factor k equals the ratio of either pair of including sides: k = AB/DE = AC/DF. Once similarity is established, the third pair (BC/EF) also has this same ratio.

Q8. Is SAS similarity important for CBSE exams?

Yes. CBSE Class 10 exams test SAS similarity in proof questions and numerical problems carrying 2–4 marks. Show the equal angle and both proportional side ratios clearly.

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