Triangles Class 10 Maths Notes

Similarity is about shape alone. A triangle and a shrunken copy of it are similar even though their sizes differ, and a triangle and its mirror image are similar because a reflection does not change any angle. Two equilateral triangles of any sizes are always similar, which is the simplest example you will ever be asked to recognise.

Before you write any ratio you must settle which vertex corresponds to which. The rule is positional: the first letter in one name goes with the first letter in the other, the second with the second, and the third with the third. So in ∆ABC ~ ∆DEF, side AB corresponds to DE and not to EF or DF, and the angles at A, B, C correspond to the angles at D, E, F.

• The letters are read in pairs: ∆ABC ~ ∆DEF gives A ↔ D, B ↔ E and C ↔ F.
• Corresponding sides: AB ↔ DE, BC ↔ EF and CA ↔ FD.
• Corresponding angles: ∠A ↔ ∠D, ∠B ↔ ∠E and ∠C ↔ ∠F.
• Writing ∆ABC ~ ∆EDF instead swaps two letters and makes the statement false unless the triangle is isosceles, so copy the order from the question exactly.
• The symbol ~ is read as 'is similar to'. The symbol = is never used between two triangles unless they are congruent.

Many students lose marks by treating a rough resemblance as similarity. The chapter is explicit that examples and counter-examples are examinable, so learn which conditions are enough and which are not.

• Equal area does not mean similar. A rectangle 2 × 1 and a rectangle 4 × 0.5 both enclose 2 square units, but their corresponding sides are in the ratios 2 and 0.5, which are not equal, so the two rectangles are not similar.
• Equal perimeter does not mean similar either, for exactly the same reason: the sides are spread differently.
• Two sides proportional on their own is not enough. A triangle with sides 3, 4, 5 and one with sides 5, 12, 13 have one pair of equal sides but no single ratio common to all three pairs, so they are not similar.
• Three sides all proportional is enough, and three angles all equal is enough. Those are two of the three criteria.
• Any two equilateral triangles are similar, because all three angles are 60° in each.

This is the one result in the chapter you must be able to PROVE. In triangle ABC, a line DE is drawn parallel to BC so that it meets AB at D and AC at E. Then the two sides AB and AC are divided in the same ratio, so AD / DB = AE / EC.

• Use the whole-side version when it suits the data: AD / AB = AE / AC, because dividing each side into a piece and the whole gives the same ratio as dividing it into two pieces.
• The same theorem in the other direction of naming: if AD / DB = AE / EC then DE ∥ BC. That reverse reading is the converse, treated in its own section below.
• The theorem is what lets a scale drawing work — it is the reason a map of a triangle has the same shape as the triangle, and the reason a scale of 1 : 50 000 keeps every angle unchanged.

In triangle ABC, DE is drawn parallel to BC with D on AB and E on AC. If AD = 3 cm, DB = 2 cm and AE = 12 cm, find EC.

• Name the theorem first, because the mark for the reasoning comes before the arithmetic.
• Cross-multiply rather than dividing 12 by 1.5 — one line, no decimal.
• State the unit with the answer, and check: AD / DB = 3 / 2 and AE / EC = 12 / 8 = 3 / 2, so the ratio agrees.

The converse reverses the logic. You are MOTIVATED to accept, not prove, that if a line divides two sides of a triangle in the same ratio, then that line is parallel to the third side. It is the statement that lets you prove a pair of lines are parallel, which the direct theorem can never do.

In triangle ABC, D lies on AB with AD = 4 cm and DB = 6 cm, and E lies on AC with AE = 6 cm and EC = 9 cm. Show that DE ∥ BC and find AE.

• Both ratios reduce to 2 / 3, so the converse applies and DE ∥ BC.
• The theorem is not needed to prove parallelism here — the converse alone is enough, and quoting the direct theorem would be wrong.
• Had AE been unknown, the same ratio would have been used forwards: AD / DB = AE / EC gives 4 / 6 = AE / EC, and with EC = 9 the whole side AC = 15 m gives AE = 6 m.

The criteria are three ways of guaranteeing similarity. All three are stated as MOTIVATED results in the rationalised syllabus, which means you may quote them in a solution — you are not expected to reconstruct the derivation of any of them.

Each criterion is examinable on its own, so learn one clean example per criterion with the names fixed in advance.

• AAA example — In ∆ABC, ∠A = 50° and ∠B = 60°; in ∆PQR, ∠P = 50° and ∠Q = 60°. Then ∠C = 180° − 50° − 60° = 70° = ∠R, so ∆ABC ~ ∆PQR by AAA and all three pairs of sides are proportional.
• SSS example — ∆ABC has sides AB = 4 cm, BC = 6 cm, CA = 8 cm and ∆PQR has sides PQ = 6 cm, QR = 9 cm, RP = 12 cm. Since 4 / 6 = 6 / 9 = 8 / 12 = 2 / 3, the triangles are similar by SSS with AB ↔ PQ, BC ↔ QR, CA ↔ RP.
• SAS example — In ∆ABC, AB = 6 cm, AC = 8 cm and ∠A = 60°; in ∆DEF, DE = 9 cm, DF = 12 cm and ∠D = 60°. Since AB / DE = 6 / 9 = 2 / 3 and AC / DF = 8 / 12 = 2 / 3, and the included angles are equal, ∆ABC ~ ∆DEF by SAS.

The rationalised syllabus removed several results that older books still print beside this chapter. If your source carries them, treat them as history and do not attempt them in the examination.

• The statement that the areas of two similar triangles are in the ratio of the squares of their corresponding sides is DELETED for 2024-25, together with its converse. Do not compute an area ratio.
• Pythagoras' theorem and its converse are DELETED from this chapter. No question this year will ask you to find a hypotenuse or a leg through them.
• The result that the perpendicular from the right angle of a right triangle to the hypotenuse creates two similar triangles is also DELETED.
• The entire Constructions chapter is DELETED, so nothing will be asked about drawing a tangent to a circle or bisecting a line segment.
• What remains is exactly this page: similarity, the BPT, its converse and the three criteria.

The unit carries 15 of the 80 theory marks, so this chapter supports both the two-mark definition questions and the full five-mark proofs. The internal assessment is 20 marks in two halves — 10 for the pen-paper test and multiple assessment, 5 for the portfolio and 5 for the lab practical — and none of that is allocated chapter by chapter.

• State the Basic Proportionality Theorem. (A one-mark MCQ or a two-mark very short answer.)
• In ∆ABC, DE ∥ BC, AD = 4 cm and DB = 6 cm. If AE = 6 cm, find EC. (A three-mark short answer.)
• Show that if AD / DB = AE / EC then DE ∥ BC, and use it to prove a stated pair of lines parallel. (A five-mark long answer.)
• Given that ∆ABC ~ ∆PQR, find the remaining sides when two sides and one corresponding side are given. (A five-mark or a case-study question.)
• State any two criteria for similarity of triangles, with a diagram. (An assertion-reason or a two-mark question.)
• A case-study question on a scale drawing or a map typically asks two sub-parts: one to compute a real distance from the scale and one to prove two lengths are equal.