CBSE Class 10 Maths, Chapter 6: Triangles

A Complete Explainer with Theorems, Q&A, and Interactive Notes

From Same Size to Same Shape

In Class 9, you learned about congruent figures. These figures have the same shape and the same size. Think of them as exact copies.

This chapter introduces a new idea: similar figures. These figures have the same shape, but not necessarily the same size. A toy car and a real car of the same model are similar.

This change from “equality” to “proportionality” is the main new idea in this chapter.

Congruence vs. Similarity

Feature	Congruence (Class 9)	Similarity (Class 10)
Definition	Same shape and same size	Same shape, not necessarily same size
Symbol	≅	~
Angles	Corresponding angles are equal.	Corresponding angles are equal.
Sides	Corresponding sides are equal.	Corresponding sides are in the same ratio (proportional).

What is Similarity Used For? The Power of Indirect Measurement

How have humans measured the height of Mount Everest or the distance to the moon? It wasn’t done with a measuring tape. These massive distances were calculated using the principles of similarity.

This idea is called indirect measurement. By creating a small, similar triangle that we *can* measure, we can set up proportions to find the lengths of a much larger, inaccessible triangle (like the one formed by you, the base of a mountain, and its peak).

The lamp-post problem you will see later is a simple version of the exact same logic used to measure mountains and planets.

Visualizing Similar Figures

Remember, “similar” means same shape, but size can be different. This means all corresponding angles are equal and all corresponding side lengths are proportional.

Always Similar

These shapes are always similar to other shapes of the same type.

All Circles are Similar

All Squares are Similar

Not Always Similar

These shapes are not always similar, as their angles can differ.

Trapezoids are Not Always Similar

Isosceles Triangles are Not Always Similar

The Foundation: Basic Proportionality Theorem (BPT)

The entire chapter is built on one main theorem, also known as Thales Theorem. It connects parallel lines to proportional sides.

Theorem 6.1: Basic Proportionality Theorem (BPT)

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.

Syllabus Status: Proof Required. This is the only major proof in this chapter that you must know for the examination.

Understanding the Proof (Area-Based)

Given: A triangle ABC where line DE is parallel to BC (DE || BC).

To Prove: (AD / DB) = (AE / EC)

Construction: Join BE and CD. Draw DM ⊥ AC and EN ⊥ AB.
Area Ratios (Left Side): We know, Area = (1/2) x base x height.
(ar(ADE) / ar(BDE)) = ((1/2) x AD x EN) / ((1/2) x DB x EN) = (AD / DB) (Eq. 1)
Area Ratios (Right Side):
(ar(ADE) / ar(DEC)) = ((1/2) x AE x DM) / ((1/2) x EC x DM) = (AE / EC) (Eq. 2)
The Key Insight (from Class 9): Triangles Triangle BDE and Triangle DEC are on the same base DE and between the same parallel lines DE and BC.
Therefore, ar(BDE) = ar(DEC). (Eq. 3)
Conclusion: From equations (1), (2), and (3), we see the denominators ar(BDE) and ar(DEC) are equal, and the numerators ar(ADE) are the same.
This means the ratios must be equal: (AD / DB) = (AE / EC). Hence Proved.

Theorem 6.2: Converse of BPT

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

Syllabus Status: Motivation Only. You must know how to *use* this theorem to prove lines are parallel, but you are not required to prove the theorem itself.

Interactive BPT Visualizer

Move the slider to change the position of the parallel line DE. Watch how the ratios of the sides change, but always remain equal to each other.

Move line DE (Slider)

Ratio 1 (Left): (AD / DB)

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Ratio 2 (Right): (AE / EC)

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The Similarity Toolkit: Proving Triangles are Similar

How do we prove two triangles are similar without checking all 6 conditions (3 angles, 3 side ratios)? We can use these three “shortcuts.”

Syllabus Status: Motivation Only. You must be an expert at *using* these three criteria to solve problems. The proofs for them are not required for exams.

1. AAA (or AA) Similarity

If two angles of one triangle are equal to two corresponding angles of another triangle, the two triangles are similar.

(We only need two angles (AA), because the third angle will automatically be equal due to the 180 degrees angle sum property).

2. SSS Similarity

If the corresponding sides of two triangles are in the same ratio (proportional), then the two triangles are similar.

Example: (AB / DE) = (BC / EF) = (AC / DF)

3. SAS Similarity

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

Example: Angle A = Angle D and (AB / DE) = (AC / DF)

An Additional Criterion: RHS Similarity

Beyond the three main criteria (AA, SSS, SAS), the textbook provides a special note for a criterion specific to right-angled triangles. This is sometimes called RHS (Right-angle, Hypotenuse, Side) Similarity.

RHS Similarity Criterion

Statement: If in two right triangles, the hypotenuse and one side of one triangle are proportional to the hypotenuse and one side of the other triangle, then the two triangles are similar.

This is a useful shortcut because you don’t need to check all three side ratios (SSS) or an included angle (SAS). You only need to confirm:

Both triangles have a right angle (90 degrees).
The ratio: (Hypotenuse of Tri 1) / (Hypotenuse of Tri 2)
Is equal to the ratio: (One Side of Tri 1) / (Corresponding Side of Tri 2)

If these two ratios are equal, the triangles are similar. This criterion is not in the main list of theorems but is provided in the textbook as a helpful note for solving problems.

Application: Solving Problems with Similarity

Similarity is most powerful for “indirect measurement,” finding heights or distances you cannot measure directly.

Solved Example: The Lamp-Post Problem

Problem: 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.

Step-by-Step Solution:

Draw and Label: Let AB be the lamp-post (3.6 m) and CD be the girl (0.9 m). Let DE be the shadow (x meters).
Find Distance: The girl walks for 4 seconds at 1.2 m/s.
Distance BD = Speed x Time = 1.2 x 4 = 4.8 m.
Identify Triangles: We have two triangles: Triangle ABE (large) and Triangle CDE (small).
Prove Similarity (AA):
- Angle E = Angle E (Common angle)
- Angle B = Angle D (Both are 90 degrees, assuming lamp and girl stand vertically)
Therefore, by AA Similarity, Triangle ABE ~ Triangle CDE.
Set up Proportions: Since the triangles are similar, the ratio of their corresponding sides is equal.
(AB / CD) = (BE / DE)
Substitute Values: We know BE = BD + DE = 4.8 + x.
(3.6 / 0.9) = (4.8 + x) / x
Solve for x:
4 = (4.8 + x) / x
4x = 4.8 + x
3x = 4.8
x = 1.6

Answer: The length of her shadow after 4 seconds is 1.6 meters.

How to Approach a Triangles Problem: A Flowchart

This chapter has many theorems. Use this flowchart to help decide which theorem or criterion to apply based on the information given in a problem. You can scroll horizontally on the chart if it doesn’t fit your screen.

Quick Q&A (Check Your Understanding)

Test your knowledge. Try to answer first, then check the hint or the full answer.

Q1: Are all equilateral triangles similar? Are all isosceles triangles similar?

Think about the two conditions for similarity: all corresponding angles must be equal, and all corresponding sides must be in ratio.

Yes, all equilateral triangles are similar. All their angles are always 60 degrees (so AAA is met). If two equilateral triangles have sides $a$ and $b$, the ratio of sides is $\frac{a}{b}$ for all three sides (so SSS is also met).

No, not all isosceles triangles are similar. One isosceles triangle could have angles 50 degrees, 50 degrees, 80 degrees and another could have 70 degrees, 70 degrees, 40 degrees. The angles are not equal, so they are not similar.

Q2: In Triangle ABC and Triangle DEF, (AB / DE) = (BC / FD). For them to be similar by SAS, which angles must be equal?

SAS requires the angle to be *included* between the two proportional sides. Look at the numerators (ABC) and denominators (DEF) separately.

Angle B = Angle D

For Triangle ABC, the two sides are AB and BC. The included angle is Angle B.

For Triangle DEF, the two sides are DE and FD. The included angle is Angle D.

Therefore, the condition for SAS similarity would be Angle B = Angle D.

Q3: If Triangle ABC ~ Triangle QRP, what is the ratio of sides equal to (AC / QP)?

The order of the letters in the similarity statement Triangle ABC ~ Triangle QRP is very important. Match the vertices.

The correspondence is: A <-> Q, B <-> R, C <-> P.

The ratio (AC / QP) matches the first and third letters of each triangle. This ratio will be equal to the ratios of the other corresponding sides:

(AB / QR) (first two letters) and (BC / RP) (last two letters).

So, (AC / QP) = (AB / QR) = (BC / RP).

Beyond the Syllabus: Area & Pythagoras Proofs

The 2024-25 syllabus has removed the *proofs* for several theorems. The focus is now on *application*. These theorems are still very important for solving problems.

Theorem 6.6: Area of Similar Triangles

Statement: The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

If Triangle ABC ~ Triangle DEF, then (ar(ABC) / ar(DEF)) = (AB / DE)² = (BC / EF)² = (AC / DF)²

Common Mistake: Students often forget to square the side ratio. If the side ratio is 1:2, the area ratio is 1²:2², which is 1:4.

Interactive Area Theorem Visualizer

Move the slider to scale Triangle DEF. Notice how the area ratio grows much faster than the side ratio.

Scale Factor (k)

Side Ratio (k)

1.50

Area Ratio (k²)

2.25

Theorems 6.8 & 6.9: Pythagoras Theorem and its Converse

The proof for the Pythagoras Theorem (a² + b² = c²) and its converse are also no longer required for examination. The proof of Pythagoras (Th 6.8) is actually a beautiful application of triangle similarity.

Proof of Pythagoras Theorem (using Similarity)

This proof is not required for exams, but it shows how similarity is the true foundation for the Pythagoras theorem.

Given: A right triangle ABC, right-angled at B.

To Prove: AC² = AB² + BC²

Construction: Draw a perpendicular line BD from B to the hypotenuse AC. (BD ⊥ AC).

Step 1 (Small Left Triangle vs. Whole Triangle): Compare Triangle ADB and Triangle ABC.
- Angle A = Angle A (Common)
- Angle ADB = Angle ABC (Both are 90 degrees)
Therefore, by AA Similarity, Triangle ADB ~ Triangle ABC.
Step 2 (Get Ratios from Step 1): Since they are similar, their sides are proportional.
(AD / AB) = (AB / AC)
Cross-multiply to get: AB² = AD x AC (Eq. 1)
Step 3 (Small Right Triangle vs. Whole Triangle): Compare Triangle BDC and Triangle ABC.
- Angle C = Angle C (Common)
- Angle BDC = Angle ABC (Both are 90 degrees)
Therefore, by AA Similarity, Triangle BDC ~ Triangle ABC.
Step 4 (Get Ratios from Step 3): Since they are similar, their sides are proportional.
(CD / BC) = (BC / AC)
Cross-multiply to get: BC² = CD x AC (Eq. 2)
Step 5 (Add the Equations): Add Equation 1 and Equation 2:
AB² + BC² = (AD x AC) + (CD x AC)
Step 6 (Factor and Conclude): Factor out AC from the right side:
AB² + BC² = AC x (AD + CD)
From the diagram, we know AD + CD = AC. Substitute this:
AB² + BC² = AC x (AC)
AB² + BC² = AC². Hence Proved.

Chapter 6: Master Theorem Summary (CBSE 2024-25)

Theorem	Common Name	Statement	Syllabus Status
Th 6.1	BPT (Thales Theorem)	If DE \|\| BC, then (AD / DB) = (AE / EC)	Proof Required
Th 6.2	Converse of BPT	If (AD / DB) = (AE / EC), then DE \|\| BC	Motivation Only
Th 6.3	AA (or AAA) Similarity	If Angle A = Angle D, Angle B = Angle E, then Triangle ABC ~ Triangle DEF	Motivation Only
Th 6.4	SSS Similarity	If (AB / DE) = (BC / EF) = (AC / DF), then Triangle ABC ~ Triangle DEF	Motivation Only
Th 6.5	SAS Similarity	If (AB / DE) = (AC / DF) and Angle A = Angle D, then Triangle ABC ~ Triangle DEF	Motivation Only
Th 6.6	Area Theorem	(ar(ABC) / ar(DEF)) = (AB / DE)²	Proof Deleted (Application only)
Th 6.8	Pythagoras Theorem	In a right triangle, a² + b² = c²	Proof Deleted (Application only)
Th 6.9	Converse of Pythagoras	If a² + b² = c², it’s a right triangle	Proof Deleted (Application only)

Study Recommendations

1. The Pragmatic Path (For Exams)

Master the *application* of all 9 theorems.
Memorize the proof for BPT (Theorem 6.1) only.
Practice solving indirect measurement problems (like the lamp-post).
Be very careful with the Area Theorem (squaring the ratio).

2. The Mastery Path (For Deeper Learning)

Do everything in the pragmatic path.
Also, study the “deleted” proofs, especially the similarity-based proof of Pythagoras (as shown above).
Understand *how* BPT (Th 6.1) is used to prove AA (Th 6.3).
This shows how the entire chapter is one single, logical system.

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CBSE Class 10 Maths, Chapter 6: Triangles: Video, Q&A Quiz