Triangles — Class 10 Mathematics

What CBSE examines here (2026-27). Similar figures and what "similar" means precisely; the Basic Proportionality Theorem (Thales) and its converse; and the three similarity criteria — AA, SSS and SAS. The chapter runs to three exercises only: 6.1, 6.2 and 6.3. The proof of Pythagoras' theorem and the theorem on areas of similar triangles were removed from this chapter — both are in the appendix at the end, marked as background.

"Two triangles are similar if their shape is identical, even if their sizes differ — the algebra of geometry."

1. About the Chapter

Class 9 was about congruence — same shape, same size. This chapter is about similarity — same shape, size free to change. That one relaxation is what makes indirect measurement possible: you cannot lay a tape measure up a tower, but you can measure its shadow.

The chapter covers, in order:

• what similarity means for figures in general, and why "angles equal" alone is not enough
• the Basic Proportionality Theorem and its converse
• three criteria that let you certify similarity from only three pieces of data

Everything here feeds Chapter 7 (the distance formula) and Chapters 8-9 (trigonometry).

2. Similar Figures

Definition

Two figures are similar if they have the same shape, whether or not they have the same size.

All circles are similar to each other. So are all squares, and all equilateral triangles. A photograph enlarged from 35 mm to 55 mm is similar to the original — every length has grown in the ratio 35 : 55, and no angle has moved.

Similar vs congruent

• Congruent (≅): same shape and same size
• Similar (~): same shape, any size

Every congruent pair is similar. The converse fails — similar figures need not be congruent.

The definition for polygons

Two polygons with the same number of sides are similar if:

1. their corresponding angles are equal, and
2. their corresponding sides are in the same ratio.

That constant ratio is called the scale factor.

Both conditions are needed. A square and a rectangle have all four angles equal, but their sides are not in a constant ratio — not similar. A square and a rhombus have sides in a constant ratio, but their angles differ — not similar. The textbook shows both counterexamples deliberately, and Exercise 6.1 Q3 tests exactly this.

3. Similarity of Triangles

Two triangles are similar if their corresponding angles are equal and their corresponding sides are in the same ratio.

Written △ABC ~ △PQR, which means:

• ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R
• AB/PQ = BC/QR = CA/RP

Correspondence is part of the statement

The order of the letters is not decoration. Writing △ABC ~ △PQR asserts A↔P, B↔Q, C↔R. If the actual matching is A↔E, B↔D, C↔F, then you must write △ABC ~ △EDF — every ratio you build afterwards depends on it.

Triangles whose corresponding angles are all equal are called equiangular. Thales' observation about them — the ratio of any two corresponding sides in two equiangular triangles is always the same — is the historical root of this whole chapter.

4. The Basic Proportionality Theorem (Thales)

Theorem 6.1

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

In △ABC with D on AB, E on AC and DE ∥ BC:

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

The proof, in outline

This is the one theorem in the chapter you can be asked to prove outright, so it is worth knowing the mechanism rather than the words.

Join BE and CD. Drop DM ⊥ AC and EN ⊥ AB. Then, using area = ½ × base × height:

• ar(ADE)/ar(BDE) = (½·AD·EN)/(½·DB·EN) = AD/DB — the heights cancel
• ar(ADE)/ar(DEC) = (½·AE·DM)/(½·EC·DM) = AE/EC — same cancellation

Now the key step: △BDE and △DEC sit on the same base DE and between the same parallels DE and BC, so they have equal areas. The two left-hand sides above are therefore equal, and so are the right-hand sides:

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

Why the parallel condition matters. It is used exactly once — to say ar(BDE) = ar(DEC). Remove the parallel and that step collapses, and with it the theorem.

The two forms of the ratio

Both of these are true, and both are useful:

• parts to parts: AD/DB = AE/EC
• wholes to wholes: AD/AB = AE/AC

What you may not do is mix them — AD/DB = AE/AC is false. This is the most common error in the whole chapter.

Theorem 6.2 — the converse

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 the tool for proving something is parallel. Whenever a question says "show that XY ∥ ZW", the converse of BPT is almost always the intended route.

5. Criteria for Similarity of Triangles

Checking all six conditions (three angles, three ratios) would be tedious. As with congruence in Class 9, three well-chosen facts are enough.

Theorem 6.3 — AAA (and hence AA)

If in two triangles the corresponding angles are equal, then their corresponding sides are in the same ratio, and the triangles are similar.

Because the angles of a triangle sum to 180°, knowing two of them fixes the third. So the criterion is usually applied in its lighter form:

AA criterion. If two angles of one triangle equal two angles of another, the triangles are similar.

This is by far the most-used criterion — every "vertically opposite angles" and "alternate angles" configuration ends here.

Theorem 6.4 — SSS similarity

If the sides of one triangle are proportional to the sides of another, their corresponding angles are equal and the triangles are similar.

To apply it safely, sort each triangle's sides and compare smallest with smallest, largest with largest. Comparing them in the order they happen to be printed is how Exercise 6.3 Q1(iii) traps people.

Theorem 6.5 — SAS similarity

If one angle of a triangle equals one angle of another and the sides including those angles are proportional, the triangles are similar.

The word including is doing all the work. The equal angle must sit between the two sides whose ratio you know. Exercise 6.3 Q1(v) hands you an equal angle and two proportional sides arranged so that the angle is not the included one — and the answer is that no conclusion follows.

A fourth, in the fine print

The chapter closes with a Note to the Reader:

If in two right triangles the hypotenuse and one side of one are proportional to the hypotenuse and one side of the other, the triangles are similar — the RHS similarity criterion.

It is not one of the numbered theorems, but it is in the book and it does shorten some proofs.

6. Worked Examples

Example 1 — BPT, finding a length

In △ABC, DE ∥ BC with D on AB and E on AC. If AD = 3 cm, DB = 5 cm and AE = 4 cm, find AC.

• BPT: AD/DB = AE/EC → 3/5 = 4/EC
• EC = 20/3 cm
• AC = AE + EC = 4 + 20/3 = 32/3 cm (≈ 10.67 cm)

Note the last step. The theorem gives you EC, not AC — forgetting to add AE back is a standard slip.

Example 2 — the converse, proving parallel

In △PQR, E lies on PQ and F on PR, with PE = 4 cm, EQ = 4.5 cm, PF = 8 cm and FR = 9 cm. Is EF ∥ QR?

• PE/EQ = 4/4.5 = 8/9
• PF/FR = 8/9
• The ratios agree, so by the converse of BPT, EF ∥ QR.

Example 3 — AA in a vertically-opposite configuration

Two segments PR and QS cross at O, with PQ ∥ RS. Show △POQ ~ △SOR.

• ∠P = ∠S and ∠Q = ∠R (alternate angles, since PQ ∥ RS)
• ∠POQ = ∠SOR (vertically opposite)
• By AA, △POQ ~ △SOR.

Example 4 — choosing between SSS and SAS

△ABC has AB = 2, BC = 2.5, CA = 3. △QRP has QR = 4, RP = 5, PQ = 6.

• AB/QR = 2/4 = 0.5
• BC/RP = 2.5/5 = 0.5
• CA/PQ = 3/6 = 0.5
• All three agree → △ABC ~ △QRP by SSS.

Watch the correspondence: it is QRP, not PQR. Read it off the ratios, not off the order the letters were given to you.

Example 5 — indirect measurement

A vertical pole 6 m tall casts a 4 m shadow. At the same moment a tower casts a 28 m shadow. How tall is the tower?

• The sun's rays hit both at the same angle, and both stand vertical, so the two right triangles are similar by AA.
• height/shadow is therefore the same for both: 6/4 = h/28
• h = 6 × 28 / 4 = 42 m

This is the whole point of the chapter — a length nobody could measure directly, obtained from two lengths anyone can.

7. Common Mistakes

1. Getting the correspondence backwards

   • △ABC ~ △PQR means A↔P, B↔Q, C↔R. Every ratio downstream inherits the error.

2. Mixing part-ratios with whole-ratios in BPT

   • AD/DB = AE/EC ✓ and AD/AB = AE/AC ✓, but AD/DB = AE/AC ✗.

3. Applying SAS with a non-included angle

   • The equal angle must lie between the two sides whose ratio you know.

4. Comparing sides in printed order for SSS

   • Sort first. Smallest with smallest.

5. Assuming equal angles make any two polygons similar

   • True for triangles. False in general — a square and a rectangle are the counterexample.

6. Stopping one step early

   • BPT gives you the segment; the question often wants the whole side. Add the pieces back.

8. Real-World Applications

Surveying and architecture

Heights of towers, trees and mountains come from shadow lengths and similar triangles. The same principle, mechanised, is what a theodolite does.

Maps and scale models

A map's scale is a similarity ratio. Every ground distance is the map distance multiplied by that one number, and every angle is preserved — which is why bearings taken off a map are usable in the field.

Photography and optics

Enlarging a negative multiplies every length by the same factor and moves no angle. The textbook opens with exactly this example.

9. Indian Heritage

• Baudhayana (~800 BCE): the Sulba Sutras state the relation between the sides of a right triangle, centuries before Pythagoras
• Aryabhata (5th century): systematic triangle computations
• Brahmagupta (7th century): cyclic quadrilaterals, extending triangle results
• Bhaskara II (12th century): area formulas and geometric proofs

10. Conclusion

Three exercises, three criteria, one theorem worth proving from memory. If you can do these four things you have the chapter:

• state and prove the Basic Proportionality Theorem
• use its converse to prove lines parallel
• pick the right criterion — AA, SSS or SAS — from the data you are given
• write the correspondence in the right order every single time

Similarity is how geometry measures what it cannot reach.

Appendix — beyond the current syllabus

Not examinable in CBSE 2026-27. Two results were removed from this chapter when the syllabus was rationalised, along with their exercises (old Exercise 6.4 and old Exercise 6.5). They are kept here because guidebooks, older question papers and most solution websites still lead with them, and students reasonably wonder whether they have missed something. You have not — the current chapter ends at Exercise 6.3.

Areas of similar triangles

The ratio of the areas of two similar triangles equals the ratio of the squares of their corresponding sides.

If △ABC ~ △PQR with sides in the ratio k : 1, then

$\frac{\text{ar}(△ABC)}{\text{ar}(△PQR)}=k^2$

The reason is quick: similarity scales the base by k, and it scales the corresponding height by k as well, so the product ½ × base × height scales by k².

Example. Two similar triangles have sides in the ratio 2 : 3. Their areas are in the ratio 4 : 9. If the smaller has area 24 cm², the larger has 24 × 9/4 = 54 cm².

Pythagoras' theorem and its similarity proof

In a right-angled triangle, the square on the hypotenuse equals the sum of the squares on the other two sides. For △ABC right-angled at B: AC² = AB² + BC².

The proof that used to live here is a neat application of similarity. Drop the perpendicular BD from the right angle B onto the hypotenuse AC. The perpendicular splits the triangle into two smaller ones, each similar to the original:

• △ADB ~ △ABC, which gives AB² = AD × AC
• △BDC ~ △ABC, which gives BC² = DC × AC

Add them:

AB² + BC² = AC × (AD + DC) = AC × AC = AC²

The converse — if the square on one side equals the sum of the squares on the other two, the angle opposite that side is a right angle — went with it.

Pythagorean triples, still worth having by heart because they save time in Chapters 7, 8 and 9: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25), (9, 40, 41), (20, 21, 29).

You still use Pythagoras in Class 10. Only its proof and its dedicated exercise left this chapter. The theorem itself is behind the distance formula in Chapter 7 and is used freely throughout Chapters 8 and 9. The chapter's own introduction still promises "a simple proof of Pythagoras Theorem learnt earlier" — a line the editors left behind when the section was cut.