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Andhra Pradesh (BIEAP)Class 10 Mathematics★ AP Board (BSEAP) SSC Mathematics · ~6 marks · Board: AP Board

Similar Triangles

SIMILAR TRIANGLES — criteria (AA, SAS, SSS), Basic Proportionality Theorem (Thales), area ratios, and Pythagoras Theorem with its converse. AP Board Class 10 Mathematics.

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By the end of this chapter you'll be able to…

1Apply AA, SAS, and SSS criteria to determine if two triangles are similar
2State and apply the Basic Proportionality Theorem (Thales Theorem) and its converse
3Use the ratio of areas theorem: Area₁/Area₂ = (side₁/side₂)²
4Apply Pythagoras Theorem and its converse to solve problems
5Write formal proofs for similarity-based theorems

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Why this chapter matters

Similar Triangles is a GEOMETRY chapter that connects to proofs, Pythagoras theorem, and real-world applications. The Basic Proportionality Theorem (Thales) is a standard 4-mark proof question in AP SSC. The area ratio theorem (ratio of areas = square of ratio of sides) is tested in application questions. Pythagoras theorem and its converse are required for all triangle problems. This chapter is important for both Class 10 and Class 11–12 geometry.

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Similar Triangles

"Similar triangles have the SAME shape — but not necessarily the SAME size."

Similarity Criteria

Criterion	What Must Be Equal
AA	Two angles
SAS	Two sides PROPORTIONAL + included angle EQUAL
SSS	Three sides PROPORTIONAL

Basic Proportionality Theorem (Thales)

A line parallel to one side of a triangle divides the other two sides in the SAME RATIO. Converse also holds.

Ratio of Areas of Similar Triangles = (Ratio of corresponding sides)².

Pythagoras Theorem: In a right triangle, hypotenuse² = sum of squares of other two sides. Converse: If a² + b² = c², the triangle is RIGHT.

Common Mistakes: Applying similarity without checking the correct CORRESPONDENCE of vertices.

Detailed Similarity Criteria

AA (Angle-Angle) Similarity

If TWO angles of one triangle are EQUAL to two angles of another triangle, the triangles are SIMILAR.

Proof Outline: If two angles are equal, the THIRD angle MUST be equal (sum of triangle = 180°). This means all three angles match, forcing the sides to be proportional.

Example 1: In △ABC and △DEF, ∠A = 50°, ∠B = 70°, ∠D = 50°, ∠E = 70°. Are the triangles similar? Yes! By AA criterion, ∠A = ∠D and ∠B = ∠E, so △ABC ∼ △DEF. 'You don't need to check the third angle — it will AUTOMATICALLY match.'

SSS (Side-Side-Side) Similarity

If the THREE sides of one triangle are PROPORTIONAL to the three sides of another, the triangles are SIMILAR.

AB/DE = BC/EF = AC/DF = k (scale factor)

Example 2: △ABC has sides 3, 4, 5. △DEF has sides 6, 8, 10. 3/6 = 4/8 = 5/10 = 1/2 → All ratios are EQUAL → △ABC ∼ △DEF ✓ 'SSS similarity requires ALL THREE ratios to be equal. If even one ratio differs, the triangles are NOT similar.'

SAS (Side-Angle-Side) Similarity

If TWO sides are PROPORTIONAL and the INCLUDED angle is EQUAL, the triangles are SIMILAR.

AB/DE = AC/DF and ∠A = ∠D → △ABC ∼ △DEF

'In SAS similarity, the angle MUST be BETWEEN the two proportional sides. This is CRITICAL.'

Example 3: In △PQR, PQ = 4, QR = 6, ∠Q = 60°. In △XYZ, XY = 6, YZ = 9, ∠Y = 60°. PQ/XY = 4/6 = 2/3, QR/YZ = 6/9 = 2/3, ∠Q = ∠Y = 60° By SAS criterion, △PQR ∼ △XYZ ✓

Basic Proportionality Theorem (Thales Theorem) — Proof

Statement: A line drawn PARALLEL to one side of a triangle divides the other two sides in the SAME RATIO.

Given: In △ABC, DE ∥ BC, where D is on AB and E is on AC. To Prove: AD/DB = AE/EC

Construction: Join D to C and E to B. Draw EF ⟂ AB and DG ⟂ AC.

Proof: ar(△ADE) = ½ × AD × EF (area = ½ × base × height) ar(△BDE) = ½ × DB × EF (same height — triangles share vertex E) So, ar(△ADE)/ar(△BDE) = AD/DB ... (i)

Similarly: ar(△ADE) = ½ × AE × DG ar(△CDE) = ½ × EC × DG So, ar(△ADE)/ar(△CDE) = AE/EC ... (ii)

But △BDE and △CDE lie on the SAME base DE and BETWEEN the SAME parallels DE and BC. Therefore, ar(△BDE) = ar(△CDE) ... (iii)

From (i), (ii), and (iii): AD/DB = AE/EC ✓

Converse: If a line divides two sides of a triangle in the SAME ratio, it is PARALLEL to the third side.

Ratio of 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 △ABC ∼ △DEF, then: ar(△ABC)/ar(△DEF) = (AB/DE)² = (BC/EF)² = (AC/DF)²

Example 4: The sides of two similar triangles are in the ratio 3:5. If the area of the smaller triangle is 36 cm², find the area of the larger triangle.

(3/5)² = 9/25 = 36/Area of larger Area of larger = 36 × 25/9 = 100 cm² ✓

'The area ratio is the SQUARE of the side ratio. If sides double, area QUADRUPLES.'

Proof Summary

△ABC ∼ △DEF → ∠A = ∠D Draw altitudes AP and DQ from A and D to BC and EF. ar(△ABC)/ar(△DEF) = (½ × BC × AP)/(½ × EF × DQ) = (BC/EF) × (AP/DQ) Since the triangles are similar, AP/DQ = AB/DE = BC/EF = AC/DF Therefore, ar(△ABC)/ar(△DEF) = (BC/EF)² ✓

Pythagoras Theorem — Proof Using Similar Triangles

Theorem: In a RIGHT triangle, the square of the HYPOTENUSE equals the sum of squares of the other two sides.

Given: Right △ABC with ∠B = 90°. To Prove: AC² = AB² + BC²

Construction: Draw BD ⟂ AC.

Proof: In △ADB and △ABC: ∠A = ∠A (common), ∠ADB = ∠ABC = 90° △ADB ∼ △ABC (AA criterion) AB/AC = AD/AB → AB² = AD × AC ... (i)

In △BDC and △ABC: ∠C = ∠C (common), ∠BDC = ∠ABC = 90° △BDC ∼ △ABC (AA criterion) BC/AC = DC/BC → BC² = DC × AC ... (ii)

Adding (i) and (ii): AB² + BC² = AD × AC + DC × AC = AC(AD + DC) = AC × AC = AC² ✓

Converse: If in a triangle, the square of one side equals the sum of squares of the other two sides, the triangle is RIGHT-ANGLED.

Worked Examples — Similar Triangles

Example 5: In △ABC, DE ∥ BC. If AD = 3 cm, DB = 6 cm, and AE = 4 cm, find EC. By BPT: AD/DB = AE/EC 3/6 = 4/EC → 3 × EC = 24 → EC = 8 cm ✓

Example 6: A vertical pole of height 6 m casts a shadow of 4 m. At the same time, a building casts a shadow of 20 m. Find the height of the building.

Since the sun's rays make the SAME angle with the ground: Height of pole/Shadow of pole = Height of building/Shadow of building 6/4 = H/20 → H = 30 m ✓

'Shadow problems use similarity — the sun's angle is the SAME for both objects at the same time.'

Example 7: In △ABC, D and E are points on AB and AC such that AD/DB = AE/EC = 1/2. If BC = 12 cm, find DE. By converse of BPT, DE ∥ BC. △ADE ∼ △ABC (by AA, since DE ∥ BC) AD/AB = DE/BC → AD/(AD + DB) = DE/BC AD/3AD = 1/3 = DE/12 → DE = 4 cm ✓

Common Mistakes — Detailed

Mistake	Correct Approach
Using AA when only ONE angle is equal	AA requires TWO angles to be equal
Confusing SAS similarity with SAS congruence	SAS similarity needs PROPORTIONALITY, not equality, of two sides
Writing similarity statement in wrong order	△ABC ∼ △DEF means A↔D, B↔E, C↔F. Write CORRESPONDING vertices.
Forgetting to square the ratio for areas	Area ratio = (side ratio)², NOT side ratio
Applying BPT with DE not parallel to BC	BPT works ONLY when DE ∥ BC
Mixing BPT with similarity	BPT gives proportional segments. Similarity gives proportional sides. Used DIFFERENTLY.

AP SSC Board Exam Focus

Topic	Marks	Frequency
AA similarity criterion	3-4	Very High
BPT and its converse	4	Very High
Ratio of areas	3	High
Pythagoras theorem	4	Very High
Shadow/height word problems	4	High
Mixed similarity problems	5	High

Self-Test Questions

In △ABC, ∠A = 60°, ∠B = 50°. In △DEF, ∠D = 60°, ∠E = 50°. Is △ABC ∼ △DEF? By which criterion?
A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from the wall.
In △ABC, DE ∥ BC. AD = 2 cm, DB = 4 cm, DE = 3 cm. Find BC.
The areas of two similar triangles are 64 cm² and 121 cm². Find the ratio of their corresponding sides.
In a right triangle, the hypotenuse is 13 cm and one side is 5 cm. Find the third side.
A girl of height 120 cm casts a shadow of 180 cm. At the same time, a tree casts a shadow of 12 m. Find the height of the tree.
In △ABC, D and E are points on AB and AC such that AD = 3 cm, DB = 5 cm, AE = 4.5 cm, EC = 7.5 cm. Is DE ∥ BC? Justify.
Prove that the sum of squares of the diagonals of a rhombus equals the sum of squares of its sides.

Answers: 1) Yes, AA criterion, 2) 6 m, 3) 9 cm, 4) 8:11, 5) 12 cm, 6) 8 m, 7) Yes, AD/DB = AE/EC = 3/5, 8) Use Pythagoras on the four right triangles formed by diagonals.

Key formulas & results

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Similar Triangles

SIMILARITY CRITERIA: AA (two angles equal), SAS (two sides proportional + included angle equal), SSS (three sides proportional). BASIC PROPORTIONALITY THEOREM (Thales): If DE ∥ BC in ΔABC, then AD/DB = AE/EC. CONVERSE: If AD/DB = AE/EC, then DE ∥ BC. AREA RATIO: If ΔABC ~ ΔPQR, then Area(ΔABC)/Area(ΔPQR) = (AB/PQ)² = (BC/QR)² = (AC/PR)². PYTHAGORAS: In right ΔABC (right angle at B): AC² = AB² + BC². CONVERSE: If AC² = AB² + BC², then ∠B = 90°.

PROOF STRUCTURE for Thales Theorem: GIVEN: DE ∥ BC. TO PROVE: AD/DB = AE/EC. CONSTRUCTION: Draw DM ⊥ AE and EN ⊥ AD. Join BE and CD. PROOF: Area(ΔADE)/Area(ΔBDE) = AD/DB [same height from E]. Area(ΔADE)/Area(ΔCDE) = AE/EC [same height from D]. Since ΔBDE and ΔCDE have the same base DE and lie between the same parallels: Area(ΔBDE) = Area(ΔCDE). Therefore AD/DB = AE/EC. AP SSC expects this proof written out completely.

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Common mistakes & fixes

These are the exact errors that cost students marks in board exams. Read them once, save yourself the trouble.

WATCH OUT

✗ Writing the wrong correspondence of vertices when applying similarity

✓ When we say ΔABC ~ ΔPQR, the ORDER of letters matters: A corresponds to P, B to Q, C to R. So AB/PQ = BC/QR = AC/PR (NOT AB/PR or BC/QP). When writing a similarity statement, always MATCH angles first (the vertex with the equal angle gets the corresponding position). Incorrect correspondence → wrong ratios → wrong answer.

Practice problems

Work through this chapter's problems as a readiness check — reveal each solution, mark yourself honestly, and get your gap report at the end.

Readiness check

Are you exam-ready for Similar Triangles?

1 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

1 questions~2 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•SIMILAR vs CONGRUENT: Similar triangles (∼) have same SHAPE but different size — all angles equal, sides proportional. Congruent triangles (≅) have same shape AND size — all angles AND all sides equal. Similarity is a weaker condition than congruence.
•THREE CRITERIA: AA (two angles equal → third angle automatically equal). SAS (one angle equal AND the sides enclosing it proportional). SSS (all three sides proportional). For AA, you need only 2 angles — the third is determined by angle sum = 180°.
•BASIC PROPORTIONALITY THEOREM (THALES): If a line is drawn parallel to one side of a triangle intersecting the other two sides, it divides them in the same ratio. DE ∥ BC in ΔABC → AD/DB = AE/EC. Converse: if AD/DB = AE/EC, then DE ∥ BC.
•AREA RATIO THEOREM: If ΔABC ∼ ΔPQR, then Area(ΔABC)/Area(ΔPQR) = (AB/PQ)² = (BC/QR)² = (CA/RP)². The ratio of areas = SQUARE of the ratio of corresponding sides. Side ratio 2:3 → area ratio 4:9.
•PYTHAGORAS THEOREM: In a right triangle with hypotenuse c and legs a, b: c² = a² + b². CONVERSE: If c² = a² + b², then the triangle has a right angle opposite side c. Common Pythagorean triples: (3,4,5), (5,12,13), (8,15,17), (7,24,25). Multiples also work: (6,8,10), (9,12,15).
•PROOF STRUCTURE — THALES THEOREM: Draw DM ⊥ AE and EN ⊥ AD. Use fact that triangles on same base with same height have equal area. Area(ΔADE)/Area(ΔBDE) = AD/DB. Area(ΔADE)/Area(ΔCDE) = AE/EC. Area(ΔBDE) = Area(ΔCDE) [same base DE, same parallels]. Therefore AD/DB = AE/EC.
•VERTEX CORRESPONDENCE IS CRITICAL: When writing ΔABC ∼ ΔPQR, the letters MUST be in correspondence order: ∠A=∠P, ∠B=∠Q, ∠C=∠R. Therefore: AB/PQ = BC/QR = CA/RP. Writing the wrong correspondence leads to wrong ratios and wrong answers.
•APPLICATIONS: A person 1.6 m tall casts a shadow 4 m long. A nearby tree casts a shadow 20 m long. The person and tree are similar to their shadows → height of tree / 1.6 = 20 / 4 → height = 8 m. This is BPT and AA similarity in real life.
•MEDIAN AND ALTITUDE RATIOS: In similar triangles, the ratio of medians = ratio of altitudes = ratio of angle bisectors = ratio of corresponding sides. If sides are in ratio k, all linear measurements (perimeters, altitudes, medians) are in ratio k. Areas in ratio k².
•COORDINATE GEOMETRY CONNECTION: If triangles are placed on coordinate axes, the distance formula and slope can verify whether sides are proportional (SSS similarity) or whether angles are equal (slope determines angle, so equal slopes verify AA similarity). This links similar triangles to Class 10 coordinate geometry.

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Where this shows up in the real world

This chapter isn't just an exam topic — it lives in the world around you.

Measuring inaccessible heights using shadows

Architects, engineers, and surveyors regularly measure the heights of buildings or natural features they cannot climb. The simplest method: measure the length of the shadow of the object at the same time as measuring the shadow of a known-height pole. The triangles formed (object + shadow + sun ray, pole + shadow + sun ray) are similar because the sun's rays are parallel. This is direct real-world application of Thales' Theorem — used daily in civil engineering and land surveying.

Camera optics and photography

A camera lens creates an inverted image on the sensor that is similar to the object. The object, lens, and image form two similar triangles (with the lens as the dividing point). The ratio of image height to object height equals the ratio of image distance to object distance — this is the lens equation (1/f = 1/v + 1/u from the optics chapter). Lens manufacturers design focal lengths using similar triangle geometry. Zooming in doubles the focal length, doubling the image size — direct consequence of scaling similar triangles.

GPS and triangulation positioning

GPS satellites determine your location by measuring the time for signals to arrive from at least 4 satellites. The satellite-to-receiver distances form triangles. The device solves a system of equations (the 3D equivalent of triangulation) to find your exact position. Triangulation — finding an unknown position using angles and distances from known points — is fundamentally about similar and congruent triangles. Every navigation system from ancient maritime charts to modern GPS applies these principles.

Exam strategy

Battle-tested tips from teachers and toppers for this chapter.

Thales Theorem proof (4-mark question): write in FIVE steps: (1) Draw perpendiculars DM and EN. (2) Write Area(ΔADE)/Area(ΔBDE) = AD/DB using same height from E. (3) Write Area(ΔADE)/Area(ΔCDE) = AE/EC using same height from D. (4) State Area(ΔBDE) = Area(ΔCDE) because same base DE and between same parallels DE and BC. (5) Conclude AD/DB = AE/EC. Each step = 1 mark. Don't skip the construction step.
Area ratio problems: identify the similarity first (write ΔABC ∼ ΔPQR with correct vertex correspondence). Then apply the area ratio = (side ratio)². Square the side ratio immediately before doing any other calculation.
Pythagorean triple check: the question 'is this a right triangle?' requires computing a²+b² and checking if it equals c². Write all three squared values explicitly — don't skip the calculation even if it seems obvious.
Vertex correspondence in similarity: BEFORE writing any ratios, draw arrows connecting corresponding vertices. ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R → then AB/PQ = BC/QR = CA/RP. Wrong correspondence = wrong ratios = zero marks for that part.
BPT application problems: always restate Thales Theorem in words for 1 mark, then set up the ratio equation for 1 mark, then solve for 1 mark, then verify for 1 mark. Writing 'By Basic Proportionality Theorem' earns a method mark even if the algebra has a minor error.

Going beyond the textbook

For olympiad aspirants and curious learners — topics that build on this chapter.

Research Euclid's proof of Pythagoras theorem (Book I, Proposition 47 of Elements) — using squares constructed on each side of a right triangle and showing equal areas by congruent triangles. This is a different and more elementary proof than the similar triangles proof. Euclid's Elements is the longest-used mathematics textbook in history (2,300+ years). Research what makes a 'proof' in mathematics and why the same theorem can have hundreds of different valid proofs.
Investigate the Pythagorean theorem in non-Euclidean geometry — on the surface of a sphere, Pythagoras fails: the sum of squares of sides of a right triangle on a sphere is NOT equal to the square of the hypotenuse. The formula is modified (spherical law of cosines). Research why this matters for navigation (Earth is a sphere, so spherical trigonometry is used for long-distance flight paths and GPS calculations).
Explore Fractal geometry — shapes that are self-similar at every scale (i.e., similar to their own parts). The Sierpinski Triangle is a famous fractal: at each stage, you remove the middle triangle, leaving three similar triangles each with half the side length. This gives an area ratio of 3 × (1/2)² = 3/4 after each step. After infinite steps, the area approaches zero. Research the Sierpinski Triangle and Mandelbrot Set as examples of infinite self-similarity.
Research the Golden Ratio (φ = 1.618...) in geometry — the Golden Rectangle can be divided into a square and a smaller Golden Rectangle (similar to the original). The ratio of consecutive Fibonacci numbers approaches φ. The Golden Ratio appears in similar triangles in regular pentagons (the ratio of diagonal to side is exactly φ). Research how the Golden Ratio connects similar triangles, Fibonacci sequences, and aesthetics in art and architecture.

Where else this chapter is tested

CBSE board isn't the only one — other exams test this chapter too.

AP Board SSC (Class 10)High — Thales Theorem proof and Pythagoras application are near-guaranteed 4–8 mark questions in AP SSC Mathematics

JEE Main (Geometry)High — Similar triangles, Pythagoras, and trigonometry in triangles form a major portion of JEE coordinate geometry and mensuration problems

NTSE (Mathematics)High — Similar triangles, BPT, and Pythagoras are standard NTSE Stage I and II topics

AP EAPCET (Mathematics)Medium — Similar triangle concepts underlie Class 11 coordinate geometry and trigonometry which are major EAPCET topics

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Questions students ask

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Area is a TWO-DIMENSIONAL measurement — it scales in two directions simultaneously. When you scale a triangle by factor k (all sides become k times longer), the BASE increases by factor k AND the HEIGHT increases by factor k. Area = (1/2) × base × height → the new area = (1/2) × (k × base) × (k × height) = k² × original area. So area scales by k² (not k). This is why: side ratio = k → area ratio = k². Example: If similar triangles have sides in ratio 3:5, their areas are in ratio 9:25. This same principle applies to all 2D shapes — a map at scale 1:100,000 has areas that are (100,000)² = 10 billion times smaller than actual area.

This is the MIDPOINT THEOREM, a direct corollary of Thales' (BPT) converse. If D is the midpoint of AB and E is the midpoint of AC, then AD/DB = 1 (since AD = DB) and AE/EC = 1 (since AE = EC). Since AD/DB = AE/EC = 1, by the converse of BPT (Thales), DE must be parallel to BC. Additionally, you can use similarity: ΔADE ∼ ΔABC (AA, since ∠A is common), and the similarity ratio is 1/2, so DE/BC = 1/2, meaning DE is HALF the length of BC. This elegant result — midpoint line is parallel to base and half its length — is tested as a direct application of similar triangles.

STRATEGY: (1) Look for parallel lines — they create equal alternate angles or corresponding angles, giving AA similarity. (2) Look for triangles sharing a common angle — that shared angle immediately gives you one equal angle for AA. (3) Look for right triangles — any two right triangles with one more equal angle are similar by AA. (4) Check ratios of sides — if two sides of one triangle are proportional to two sides of another AND the included angles are equal, it is SAS similarity. (5) For altitude drawn from right angle vertex to hypotenuse — this ALWAYS creates two triangles similar to the original and to each other. Practice: when you spot two triangles, list all shared angles and parallel lines before writing the similarity statement.

Check: 5² + 12² = 25 + 144 = 169 = 13². Since a² + b² = c², this is a right triangle with the right angle opposite the longest side (13). It IS a valid Pythagorean triple. Quick method: always check the LARGEST number squared equals the sum of squares of the other two. Common AP SSC Pythagorean triples to memorise: (3,4,5) — 9+16=25 ✓. (5,12,13) — 25+144=169 ✓. (8,15,17) — 64+225=289 ✓. (7,24,25) — 49+576=625 ✓. Any multiple of these (e.g., 6,8,10) is also a Pythagorean triple.

Similar triangles are the foundation of INDIRECT MEASUREMENT — measuring things we cannot reach. (1) THALES himself used shadow ratios to measure the height of the Great Pyramid of Giza — by comparing the pyramid's shadow to the shadow of a known pole. (2) TRIGONOMETRY is built on similar triangles — all trig ratios are ratios of sides of similar right triangles (a 30-60-90 triangle of any size has the same ratios). (3) MAPS are scale models — every map uses the principle that the map triangle and the actual terrain triangle are similar, with area scaled by (map scale)². (4) PHOTOGRAPHY — camera lenses create similar triangles between object and image. (5) SURVEYING — surveyors use triangulation (measuring angles from two known points) which relies entirely on similar and congruent triangles.

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