Class 10 Maths Chapter 6: Triangles Notes & Formulas

Exam Weightage & Blueprint

Total: 7-9 Marks

This chapter falls under Unit IV: Geometry (15 marks total). As per the latest syllabus, focus is on: proof and application of the Basic Proportionality Theorem (BPT) and criteria for similarity of triangles (AAA, SSS, SAS).

Question Type	Marks	Frequency	Focus Topic
MCQ	1	High	Basic Proportionality Theorem (BPT) Applications
Short Answer	2 or 3	Medium	Similarity Criteria (AA, SAS, SSS)
Long Answer	5	High	Proof of BPT or Similarity Theorems

CBSE 2026–27 Syllabus Status: Triangles

Unit IV: Geometry • 7–9 Marks

This chapter is strictly structured according to the CBSE Rationalised Mathematics Syllabus. Focus your board preparation only on prescribed topics and avoid studying deleted sections:

In 2026–27 Board Syllabus:

Deleted / Not in Board Exam:

⏰ Last 24-Hour Checklist

Congruent vs Similar: Similar = same shape, diff size.
BPT (Thales Theorem): $$\frac{AD}{DB}$ = $\frac{AE}{EC}$$ if $DE || BC$.
Similarity Criteria: AAA, AA, SSS, SAS.

Perimeter Ratio Theorem: Ratio of perimeters = Ratio of corresponding sides, altitudes, and medians.
Shadow Problems: Use similarity of triangles formed by sun rays.

📐 Theorems & Proofs

Basic Proportionality Theorem (BPT / Thales Theorem – 5 Marks): 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.

Full Step-by-Step Board Proof of BPT (Theorem 6.1)

Given: A triangle $ABC$ in which a line parallel to side $BC$ intersects other two sides $AB$ and $AC$ at $D$ and $E$ respectively ($DE \parallel BC$).

To Prove: $\frac{AD}{DB} = \frac{AE}{EC}$.

Construction: Join $BE$ and $CD$. Draw $DM \perp AC$ and $EN \perp AB$.

Figure: Basic Proportionality Theorem Construction & Altitudes

Step 1: Area of $\triangle ADE = \frac{1}{2} \times base \times height = \frac{1}{2} \times AD \times EN$

Step 2: Area of $\triangle BDE = \frac{1}{2} \times DB \times EN \quad (height is altitude EN on extended side DB)$

Taking ratio: $\frac{ar(\triangle ADE)}{ar(\triangle BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \quad --- (1)$

Step 3: Similarly, taking base $AE$ and altitude $DM$ on $AC$:

$ar(\triangle ADE) = \frac{1}{2} \times AE \times DM$, $ar(\triangle DEC) = \frac{1}{2} \times EC \times DM$

$\frac{ar(\triangle ADE)}{ar(\triangle DEC)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \quad --- (2)$

Step 4: $\triangle BDE$ and $\triangle DEC$ are on the same base $DE$ and between the same parallels $DE$ and $BC$.

$\therefore ar(\triangle BDE) = ar(\triangle DEC) \quad --- (3)$

From (1), (2), and (3):

$$ \frac{AD}{DB} = \frac{AE}{EC} \quad [Hence Proved] $$

🎯 Important Corollaries of BPT (Used Heavily in MCQs & 2-Mark Questions):

$\frac{AD}{AB} = \frac{AE}{AC}$ (Adding 1 to both sides: $\frac{AD}{DB} + 1 = \frac{AE}{EC} + 1$)
$\frac{DB}{AB} = \frac{EC}{AC}$

Converse of BPT (Theorem 6.2): If a line divides any two sides of a triangle in the same ratio ($\frac{AD}{DB} = \frac{AE}{EC}$), then the line is parallel to the third side ($DE \parallel BC$).

Internal Angle Bisector Theorem (Competitive/Optional Edge): The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle: $\frac{BD}{DC} = \frac{AB}{AC}$.

📐 Similarity Criteria

AAA / AA

If corresponding angles are equal, triangles are similar.

SSS

If corresponding sides are proportional, triangles are similar.

SAS

One angle equal and sides including it are proportional.

Note: AA is sufficient for similarity. You don't need to prove all three angles.

⚡ High-Yield Conceptual Rules (Board MCQs & A-R):

Converse of BPT (CBPT): If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Exam Distinction: Use BPT to find lengths when lines are parallel; use Converse of BPT to prove lines are parallel.
Perimeter Ratio Theorem (High-Frequency MCQ): If $\triangle ABC \sim \triangle DEF$, then: $$\frac{\text{Perimeter}(\triangle ABC)}{\text{Perimeter}(\triangle DEF)} = \frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = \frac{\text{Altitude}_1}{\text{Altitude}_2} = \frac{\text{Median}_1}{\text{Median}_2}$$
Right Triangle Similarity Theorem (Theorem 6.7): If $BD \perp AC$ in right $\triangle ABC$ ($ngle B = 90^\circ$): $$\triangle ADB \sim \triangle BDC \sim \triangle ABC \implies BD^2 = AD \cdot CD \quad \text{and} \quad AB^2 = AD \cdot AC$$

ΔABC ∼ ΔDEF: Equiangular & Corresponding Sides are in Proportion

Solved Examples (Board Marking Scheme)

Q1. In $\triangle ABC$, $DE || BC$. If $AD=1.5$ cm, $DB=3$ cm, $AE=1$ cm, find $EC$. (2 Marks)

Step 1: State Theorem 0.5 Mark

Since $DE || BC$, by Basic Proportionality Theorem:

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

Step 2: Substitute Values 1 Mark

$\frac{1.5}{3} = \frac{1}{EC}$

$\frac{1}{2} = \frac{1}{EC}$

Step 3: Solve 0.5 Mark

$EC = 2$ cm.

Q2. Diagonals of a trapezium ABCD with $AB || DC$ intersect at O. Show that $$\frac{AO}{BO}$ = $\frac{CO}{DO}$$. (3 Marks)

Step 1: Identify Triangles 1 Mark

Consider $\triangle AOB$ and $\triangle COD$.

$\angle AOB = \angle COD$ (Vertically opposite angles)

Step 2: Alternate Angles 1 Mark

Since $AB || DC$, alternate interior angles are equal:

$\angle OAB = \angle OCD$

Step 3: Apply Similarity 1 Mark

By AA criterion, $\triangle AOB \sim \triangle COD$.

Therefore, corresponding sides are proportional:

$\frac{AO}{CO} = \frac{BO}{DO} \Rightarrow \frac{AO}{BO} = \frac{CO}{DO}$ (Rearranging terms)

Previous Year Questions (PYQs)

2023 (1 Mark): If $\triangle ABC \sim \triangle DEF$ with $AB = 3\text{ cm}, DE = 4.5\text{ cm}$, and perimeter of $\triangle ABC = 18\text{ cm}$, find the perimeter of $\triangle DEF$.
Ans: $\frac{\text{Perimeter}(\triangle ABC)}{\text{Perimeter}(\triangle DEF)} = \frac{AB}{DE} \implies \frac{18}{P} = \frac{3}{4.5} \implies P = \frac{18 \times 4.5}{3} = 27\text{ cm}$.

2020 (3 Marks): A vertical pole of length 6m casts a shadow 4m long. At the same time, a tower casts a shadow 28m long. Find height of tower.
Ans: $\triangle ABC \sim \triangle PQR$ (Sun's elevation is same). $$\frac{6}{h}$ = $\frac{4}{28}$ \Rightarrow h = 42$ m.

2019 (4 Marks): In a right-angled triangle $ABC$, right-angled at $B$, a perpendicular $BD$ is drawn to the hypotenuse $AC$. Prove that $\triangle ADB \sim \triangle ABC$ and $BD^2 = AD \cdot CD$.
Ans: In $\triangle ADB$ and $\triangle ABC$: $\angle ADB = \angle ABC = 90^\circ$ and $\angle A = \angle A$ (common) $\implies \triangle ADB \sim \triangle ABC$ (by AA). Similarly, $\triangle BDC \sim \triangle ABC$. Thus $\triangle ADB \sim \triangle BDC \implies \frac{BD}{CD} = \frac{AD}{BD} \implies BD^2 = AD \cdot CD$.

Exam Strategy & Mistake Bank

⚠️ Mistake Bank

Order of Vertices: Writing $\triangle ABC \sim \triangle DEF$ when actually $\triangle ABC \sim \triangle EDF$. Order matters!

BPT vs Similarity: Confusing $\frac{AD}{DB} = \frac{AE}{EC}$ (BPT) with $\frac{AD}{AB} = \frac{DE}{BC}$ (Similarity).

Drawing Figures: Not drawing a figure for geometry questions. It is mandatory for marks.

💡 Scoring Tips

Correspondence: Always check which angle corresponds to which before writing similarity.

Given/To Prove: Always write "Given", "To Prove", and "Construction" headings in theorem proofs.

Shadow Problems: Remember "at the same time" means the angle of elevation of the sun is equal for both objects.

Self-Assessment Mock Test (10 Marks)

Q1 (1M): All equilateral triangles are ___________ (congruent / similar).

Q2 (2M): In $\triangle PQR$, $S$ and $T$ are points on $PQ$ and $PR$ respectively such that $ST \parallel QR$. If $PS = 3$ cm, $SQ = 3$ cm, and $PT = 4$ cm, find $PR$.

Q3 (3M): 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.

Q4 (4M): State and prove the Basic Proportionality Theorem (Thales Theorem).

📝 More Solved Board Questions

Q3. In $\triangle ABC$ and $\triangle DEF$, $\angle A = \angle D$, $\angle B = \angle E$, $AB = 3$ cm, $BC = 5$ cm, and $DE = 6$ cm. Find $EF$. 2 Marks

Sol. Since two angles are equal, $\triangle ABC \sim \triangle DEF$ by AA Similarity Criterion.

Therefore, $\frac{AB}{DE} = \frac{BC}{EF}$

$\frac{3}{6} = \frac{5}{EF} \Rightarrow \frac{1}{2} = \frac{5}{EF}$

$EF = 10$ cm

Q4. E and F are points on the sides PQ and PR respectively of a $\triangle PQR$. For $PE = 3.9$ cm, $EQ = 3$ cm, $PF = 3.6$ cm and $FR = 2.4$ cm, check if $EF \parallel QR$. 3 Marks

Sol. By Converse of BPT, if ratios are equal, lines are parallel.

$\frac{PE}{EQ} = \frac{3.9}{3} = 1.3$

$\frac{PF}{FR} = \frac{3.6}{2.4} = \frac{3}{2} = 1.5$

Since $\frac{PE}{EQ} \neq \frac{PF}{FR}$, $EF$ is NOT parallel to $QR$.

Q5. In $\triangle ABC$, $DE \parallel BC$ intersecting $AB$ at $D$ and $AC$ at $E$. If $AD = x, DB = x - 2, AE = x + 2$ and $EC = x - 1$, find the value of $x$. 3 Marks (Algebraic BPT)

Sol. Since $DE \parallel BC$, by BPT: $\frac{AD}{DB} = \frac{AE}{EC}$

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

Cross-multiplying: $x(x - 1) = (x - 2)(x + 2)$

$$x^2 - x = x^2 - 4 \implies -x = -4 \implies \mathbf{x = 4}$$

Check: $AD = 4, DB = 2 \implies 4/2 = 2$; $AE = 6, EC = 3 \implies 6/3 = 2$ ✅ Verified.

🎯 Board Pattern (2018–2025): The proof of the Basic Proportionality Theorem (BPT) is one of the most frequently asked 5-mark questions in geometry. Ensure you write the "Given", "To Prove", and "Construction" parts clearly, as they carry marks even if the proof is incomplete.

CBSE Section E • 4 Marks

Case Study: Smart Streetlighting & Geometric Shadows

Scenario: During an evening walk, a student of height 90 cm (0.9 m) walks away from the base of a smart city lamp post at a steady speed of 1.2 m/s. The lamp is mounted at the top of the post at a height of 3.6 m above the ground.

Q1: State the geometric similarity criterion applicable between the triangle formed by the lamp post and the triangle formed by the student. 1 Mark

Q2: Find the distance walked by the student after 4 seconds. 1 Mark

Q3: Find the length of the student's shadow after 4 seconds. (OR) If the student continues walking and the shadow length becomes 2.4 m, how far is the student from the lamp post? 2 Marks

📋 Board Revision Checklist

✅ All congruent figures are similar, but similar figures need not be congruent
✅ Two polygons are similar if (i) corresponding angles are equal and (ii) corresponding sides are in the same ratio
✅ BPT Theorem: Parallel line divides other two sides proportionally
✅ AA Similarity: Only 2 angles equal are enough for similarity
✅ SAS Similarity: Ratio of 2 sides and the angle between them are equal
✅ SSS Similarity: Ratios of all three corresponding sides are equal
✅ Order of similarity: $\triangle ABC \sim \triangle PQR$ \implies $\angle A=\angle P$, $\angle B=\angle Q$, etc.
✅ Note: Area of Similar Triangles and Pythagoras Theorem are not included in the 2026-27 CBSE Class 10 Mathematics syllabus

💡 Exam Tip:
In similarity questions, always name the triangles in the correct order of correspondence. If $\triangle ABC \sim \triangle RQP$, then $\frac{AB}{RQ} = \frac{BC}{QP} = \frac{AC}{RP}$.

Concept Mastery Quiz 🎯

Test your readiness for the board exam.

1. If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. This is:

2. All ___________ triangles are similar.

3. In $\triangle ABC$, $DE || BC$. If $AD=3$, $DB=5$, $AE=6$, then $EC$ is:

4. If $\triangle ABC \sim \triangle PQR$ such that $\angle A = 45^^\circ$ and $\angle B = 75^^\circ$, then $\angle R$ is:

5. Two triangles are similar by SAS criterion if:

Frequently Asked Questions: Class 10 Maths Chapter 6

What is the marks weightage of Triangles in CBSE Class 10?

Triangles carries around 6 to 8 marks in Unit 4: Geometry, usually including a mandatory 3-mark or 5-mark theorem proof or theorem application problem.

Which theorem proof is compulsory in the Class 10 Triangles syllabus?

The proof of the Basic Proportionality Theorem (Thales' Theorem) is the key theorem tested in the CBSE board exam. Areas of similar triangles and Pythagoras theorem proofs are deleted from the rationalized syllabus.

What are the criteria for similarity of two triangles?

The three criteria for triangle similarity are AAA (or AA) criterion, SSS criterion, and SAS criterion.

How should I present a geometry proof to score full marks?

Strictly structure your solution into five distinct headings: Given, To Prove, Construction (if any), Figure (drawn with ruler and labelled), and Step-by-step Proof with mathematical reasons in brackets.

Chapter 6 Study Modules & Practice Resources

Complete Notes & Formulas • NCERT Solutions • Previous Year Questions • Chapter Tests & Quizzes • Practice Worksheets

Chapter 6: Triangles

Overview