What are the three criteria for similarity of triangles in CBSE Class 10?

1. **AA / AAA Similarity:** If two angles of one triangle are respectively equal to two angles of another triangle. 2. **SAS Similarity:** If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional. 3. **SSS Similarity:** If all three corresponding sides of two triangles are in the same ratio.

What is the crucial difference between Congruence ($cong$) and Similarity ($sim$)?

Congruent triangles have the exact same shape and same size (ratio of sides = 1). Similar triangles have the same shape but not necessarily the same size (corresponding angles are equal and corresponding sides are in a constant ratio k). All congruent triangles are similar, but similar triangles need not be congruent.

How do you write the similarity statement in correct corresponding vertex order?

Corresponding vertices must strictly align in the naming. If \angle A = \angle D, \angle B = \angle E, \angle C = \angle F, write \Delta ABC \sim \Delta DEF. This automatically implies \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}. Writing \Delta ABC \sim \Delta EDF is mathematically incorrect and causes ratio errors.

When should I use BPT versus Triangle Similarity to solve an unknown segment?

Use BPT when comparing segments on the divided lateral sides (e.g., \frac{AD}{DB} = \frac{AE}{EC}). If the problem involves the parallel segment itself (DE and BC), you MUST use similarity (Delta ADE \sim \Delta ABC \implies \frac{AD}{AB} = \frac{DE}{BC}), because BPT does not relate to the base.

Are Area of Similar Triangles Theorem and Pythagoras Theorem proofs in CBSE 2026–27?

No. The Ratio of Areas of Similar Triangles theorem ((\frac{\text{area}_1}{\text{area}_2} = (\frac{s_1}{s_2})^2) and the Pythagoras Theorem proof have been removed from the CBSE syllabus. The active focus is purely on BPT and AA/SAS/SSS similarity criteria.

What geometric construction must be remembered for the formal proof of BPT?

In \Delta ABC with DE \parallel BC: Join BE and CD. Draw perpendicular heights DM \perp AC and EN \perp AB. Express areas using \frac{1}{2} \times \text{base} \times \text{height}, and use the property that \Delta BDE and \Delta CDE lie on the same base DE between the same parallels DE \parallel BC.

All Revision Notes

MathematicsCh-6 15 min comprehensive revision

NCERT Class 10 Mathematics — Chapter 6

Triangles

Basic Proportionality Theorem (Thales' Theorem) and its converse, criteria for similarity of triangles (AAA/AA, SSS, SAS), and geometric riders.

Quick Key Takeaways:

Basic Proportionality Theorem (BPT / Thales Theorem): If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio:

\frac{AD}{DB} = \frac{AE}{EC}

Converse of BPT: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

Similarity Criteria: (1) AA Similarity: Two angles equal; (2) SSS Similarity: Corresponding sides are proportional; (3) SAS Similarity: One angle equal and including sides proportional.

Crucial Distinction: Congruent figures have identical shape and size; Similar figures have identical shape but proportional sizes.

Jump to Concept:1. 1. Geometric Similarity Principles & The Basic Proportionality Theorem 2. 2. Rigorous Proof of Basic Proportionality Theorem (Thales' Theorem)3. 3. Important Solved Board Examination Questions (3-Mark & 5-Mark)4. 4. CBSE Case-Study Modeling: Shadow Method for Measuring Monument Heights 5. 5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

1. Geometric Similarity Principles & The Basic Proportionality Theorem

Axioms & Foundational Theory

Comprehensive mathematical foundations, definitions, formulas, and visual conceptual model for Triangles.

• Axiomatic Definition of Similarity of Triangles

•

Two triangles

\Delta ABC

and

\Delta DEF

are similar (denoted

\Delta ABC \sim \Delta DEF

) if and only if:

•

1. Their corresponding angles are equal:

\angle A = \angle D, \angle B = \angle E, \angle C = \angle F

•

2. Their corresponding sides are in the same ratio (proportional):

\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF}

•

AA Similarity Criterion Theorem: If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar (since the third angles are automatically equal by the Angle Sum Property).

📊 Triangles: Basic Proportionality Theorem & SimilarityVisual Model

Visual schematic illustrating the Thales Theorem parallel line division, triangle altitude constructions, and AA similarity ratio rules.

2. Rigorous Proof of Basic Proportionality Theorem (Thales' Theorem)

Theorem Proofs & Derivations

Step-by-step rigorous geometric and algebraic theorem proofs and formula derivations for Triangles.

• Theorem 6.1 (BPT): Step-by-Step Formal Geometric Proof

•

Statement: If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides in the same ratio.

•

Given: A triangle

\Delta ABC

in which a line

DE \parallel BC

intersects

AB

D

and

AC

E

•

To Prove:

\frac{AD}{DB} = \frac{AE}{EC}

•

Construction: Join

BE

and

CD

. Draw

DM \perp AC

and

EN \perp AB

•

Proof:

•

\text{Area}(\Delta ADE) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AD \times EN

•

\text{Area}(\Delta BDE) = \frac{1}{2} \times DB \times EN

(since

EN

is altitude on base

AB

)

•

\frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \quad \text{--- (Equation 1)}

•

3. Similarly, taking base

AE

and

EC

with altitude

DM

•

\frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta CDE)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \quad \text{--- (Equation 2)}

•

4. Notice that

\Delta BDE

and

\Delta CDE

lie on the same base $DE$ and between the same parallel lines $DE \parallel BC$ :

•

\text{Area}(\Delta BDE) = \text{Area}(\Delta CDE) \quad \text{--- (Equation 3)}

•

5. From Equations (1), (2), and (3), the left-hand sides are equal. Hence:

\mathbf{\frac{AD}{DB} = \frac{AE}{EC}}

•

Hence Proved.

3. Important Solved Board Examination Questions (3-Mark & 5-Mark)

Topper Step Solutions

Standard CBSE board exam numericals and riders with complete step-by-step mathematical working and justifications.

• 3-Mark Standard Board Question: In $\Delta ABC$ , $DE \parallel BC$ such that $AD = x, DB = x - 2, AE = x + 2$ , and $EC = x - 1$ . Find the value of $x$ .

•

Step 1 (Applying BPT): Since

DE \parallel BC

, by Basic Proportionality Theorem:

\frac{AD}{DB} = \frac{AE}{EC}

•

Step 2 (Substituting Values):

\frac{x}{x - 2} = \frac{x + 2}{x - 1}

•

Step 3 (Cross-Multiplication):

x(x - 1) = (x - 2)(x + 2)

x^2 - x = x^2 - 4

-x = -4 \implies \mathbf{x = 4}

•

Final Boxed Answer:

\mathbf{x = 4}

• 5-Mark Heavyweight Board Problem / Rider: Prove that if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side (Converse of BPT).

•

Given: A triangle

\Delta ABC

and a line

DE

intersecting

AB

D

and

AC

E

such that

\frac{AD}{DB} = \frac{AE}{EC}

•

To Prove:

DE \parallel BC

•

Proof by Contradiction:
1. Let us assume that

DE

is not parallel to

BC

.
2. Then there must exist another line through

D

, say

DE'

, which is parallel to

BC

.
3. Since

DE' \parallel BC

, by Basic Proportionality Theorem:

\frac{AD}{DB} = \frac{AE'}{E'C} \quad \text{--- (1)}

4. But it is given that:

\frac{AD}{DB} = \frac{AE}{EC} \quad \text{--- (2)}

5. Equating (1) and (2):

\frac{AE'}{E'C} = \frac{AE}{EC}

6. Adding 1 to both sides:

\frac{AE'}{E'C} + 1 = \frac{AE}{EC} + 1 \implies \frac{AE' + E'C}{E'C} = \frac{AE + EC}{EC} \implies \frac{AC}{E'C} = \frac{AC}{EC}

7. Therefore,

E'C = EC

. This is possible only if the points

E

and

E'

coincide.
8. Hence, our assumption was false, and $DE \parallel BC$ .

•

Hence Proved.

4. CBSE Case-Study Modeling: Shadow Method for Measuring Monument Heights

Case Study Mastery (4 Marks)

Real-world mathematical modeling scenario with structured multi-part questions and full step solutions.

• Practical Application Context: Shadow Method for Measuring Monument Heights

A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.

•

Q1: Set up the similarity of triangles in this configuration.

\rightarrow

Let lamp-post be

AB = 3.6

m, girl be

CD = 90\text{ cm} = 0.9

m. After 4 seconds, distance walked

BD = 1.2 \times 4 = 4.8

m. Let shadow length

DE = x

m. In

\Delta ABE

and

\Delta CDE

\angle B = \angle D = 90^\circ

and

\angle E = \angle E

(common). By AA Similarity,

\mathbf{\Delta ABE \sim \Delta CDE}

•

Q2: Calculate the length of her shadow ( $x$ ).

\rightarrow

\frac{BE}{DE} = \frac{AB}{CD} \implies \frac{BD + DE}{DE} = \frac{3.6}{0.9} \implies \frac{4.8 + x}{x} = 4 \implies 4.8 + x = 4x \implies 3x = 4.8 \implies \mathbf{x = 1.6\text{ meters}}

•

Q3: What is the ratio of the area of $\Delta CDE$ to $\Delta ABE$ ?

\rightarrow

\left(\frac{CD}{AB}\right)^2 = \left(\frac{0.9}{3.6}\right)^2 = \left(\frac{1}{4}\right)^2 = \mathbf{\frac{1}{16}}

5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

Important Solved Board Questions

Examiner step-marking allocations, common algebraic/arithmetic deduction traps, and presentation guidelines.

• Step-by-Step Marking Rubric & Presentation Guidelines

•

1 Mark: Complete Given, To Prove, and Construction statements with labelled diagram.

•

1 Mark: Correct area equations

\text{Area}(\Delta ADE)/\text{Area}(\Delta BDE) = AD/DB

•

1 Mark: Justifying

\text{Area}(\Delta BDE) = \text{Area}(\Delta CDE)

(same base between parallel lines).

•

1 Mark: Final equivalence deduction.

• Common Calculation Traps & Verification Checklist

•

Trap 1: Forgetting to justify 'triangles on the same base between same parallels have equal areas' in BPT proof.

•

Trap 2: Unit mismatch: mixing centimeters (

90

cm) and meters (

3.6

m) without converting to common units.

•

Trap 3: Miswriting similarity order (e.g. writing

\Delta ABC \sim \Delta EFD

when vertices do not correspond).

Authentic Board Question (5 Marks)Topic: Triangles Similarity Theorems

State and prove Basic Proportionality Theorem (Thales Theorem).

Official CBSE Step-by-Step Marking Breakdown:

Step 1: Statement & Given Figure: Accurate formal statement + neatly labelled

\Delta ABC

with line

DE \parallel BC

1 Mark

Step 2: Construction: Join

BE, CD

and draw perpendiculars

DM \perp AC

and

EN \perp AB

1 Mark

Step 3: Area Ratio Expressions: Express

\frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta BDE)} = \frac{AD}{DB}

and

\frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta CDE)} = \frac{AE}{EC}

2 Marks

Step 4: Equating & Final Proof: State

\Delta BDE

and

\Delta CDE

are on same base

DE

and between same parallels

DE \parallel BC \implies \frac{AD}{DB} = \frac{AE}{EC}

1 Mark

Model Student Answer (Target: Full 5/5 Marks):

Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Given: In

\Delta ABC

, a line parallel to side

BC

intersects other two sides

AB

and

AC

D

and

E

respectively.
To Prove:

\frac{AD}{DB} = \frac{AE}{EC}

Construction: Join

BE

and

CD

. Draw

DM \perp AC

and

EN \perp AB

.
Proof:

\text{Area}(\Delta ADE) = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times AD \times EN

\text{Area}(\Delta BDE) = \frac{1}{2} \times DB \times EN

\implies \frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta BDE)} = \frac{AD}{DB}

...(1)
Similarly,

\text{Area}(\Delta ADE) = \frac{1}{2} \times AE \times DM

and

\text{Area}(\Delta CDE) = \frac{1}{2} \times EC \times DM

\implies \frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta CDE)} = \frac{AE}{EC}

...(2)
Note that

\Delta BDE

and

\Delta CDE

are on the same base

DE

and between the same parallels

BC

and

DE

.
Therefore,

\text{Area}(\Delta BDE) = \text{Area}(\Delta CDE)

...(3)
From (1), (2), and (3):
$\frac{AD}{DB} = \frac{AE}{EC}$ (Hence Proved).

Examiner Mark Deduction Traps:

•Never omit the 4-part structure: Given

\to

To Prove

\to

Construction

\to

Proof.

•Drawing the figure with a pencil and ruler is expected in board answer booklets.

High-Frequency Conceptual Doubts & FAQs

Curated answers to the most common questions asked by Class 10 students.

BPT states: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. In

\Delta ABC

, if

DE \parallel BC

, then

\frac{AD}{DB} = \frac{AE}{EC}

Triangles

1. Geometric Similarity Principles & The Basic Proportionality Theorem

2. Rigorous Proof of Basic Proportionality Theorem (Thales' Theorem)

3. Important Solved Board Examination Questions (3-Mark & 5-Mark)

4. CBSE Case-Study Modeling: Shadow Method for Measuring Monument Heights

5. CBSE Examiner Marking Scheme, Calculation Traps & Step Rubric

Official CBSE Step-by-Step Marking Breakdown:

High-Frequency Conceptual Doubts & FAQs

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