Similarity of triangles

Hold a photograph up to a projector screen and the enlarged image is a different size but unmistakably the same picture — every angle in it matches, and every length has grown by the same factor. That is precisely what it means for two triangles to be similar: same shape, angles matching one-to-one, and sides matching in a single constant ratio called the scale factor. Similarity is the second building block of the CBSE Class 10 chapter Triangles, sitting right after "similar figures" and just before the AAA, SSS and SAS criteria that let you prove similarity from partial data. Master this definition first — the correspondence notation, the angle-equality facts, and the side-ratio facts — because every criterion, every proportional-length problem, and later the Basic Proportionality Theorem and Pythagoras theorem all lean on it.

The core insight

Two triangles are similar exactly when their corresponding angles are equal and their corresponding sides are in one constant ratio, the scale factor.

The core idea

Two triangles △ABC\triangle ABC and △DEF\triangle DEF are called similar, written △ABC∼△DEF\triangle ABC \sim \triangle DEF, when both of the following hold together:

  • their corresponding angles are equal, and
  • their corresponding sides are in the same ratio (proportional).

The vertex order in the similarity statement is not decoration — it defines the correspondence: A↔DA \leftrightarrow D, B↔EB \leftrightarrow E, C↔FC \leftrightarrow F. With that correspondence fixed,

∠A=∠D,∠B=∠E,∠C=∠F\angle A=\angle D,\quad \angle B=\angle E,\quad \angle C=\angle F
ABDE=BCEF=CAFD=k\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}=k

where: kk is the scale factor, the single number every side of △DEF\triangle DEF is multiplied by to get the matching side of △ABC\triangle ABC's partner (or vice versa, depending on which way the ratio is written). If k>1k>1 the second triangle is an enlargement of the first, k<1k<1 a reduction, and k=1k=1 makes them congruent — similar and equal in size. A fact special to triangles (it fails for general polygons) is that either one of the two bullet conditions above is enough to force the other; that shortcut is exactly what the AAA/AA, SSS and SAS similarity criteria, covered separately, make precise. Here the focus is on what similarity means once it is established, and how to use it. Two habits keep the definition from tripping you up. First, corresponding sides sit opposite equal angles: the side opposite the largest angle in one triangle matches the side opposite the largest angle in the other, so you can sanity-check a proposed correspondence just by comparing which sides look "long" versus "short." Second, similar triangles usually arise embedded inside one figure — a line drawn parallel to a side, or two triangles sharing a vertex with parallel bases — and matching vertices correctly, not by eye, must happen before any ratio is written down.

Figure
Two similar triangles ABC and DEF with equal marked angles and side ratios shown by matching tick marks
Similar triangles — equal angles, sides in constant ratio k A B C / / // D E F / / // triangle ABC triangle DEF, scale factor k larger

Common traps

  • Writing the similarity statement with the wrong vertex order — △ABC∼△PQR\triangle ABC \sim \triangle PQR is not the same claim as △ABC∼△QPR\triangle ABC \sim \triangle QPR; the order fixes which angles and sides correspond, and reading ratios off the wrong pairing gives a wrong answer even with correct arithmetic.
  • Assuming any two triangles that "look alike" are similar without checking a stated criterion or given data — similarity must be established (via AAA, SAS or SSS) or given outright, never assumed from a diagram's appearance.
  • Confusing "corresponding sides are proportional" with "corresponding sides are equal" — similar triangles need not have equal sides at all; only congruent triangles do.
  • Forgetting that the third pair of angles is automatically equal once two pairs match, since angles of a triangle sum to 180∘180^\circ — trying to verify all three independently wastes time when two are enough.
Derivation AWhy the perimeters of similar triangles share the triangles' scale factor

The goal is to show, starting only from the definition of similar triangles, that the ratio of the perimeters of two similar triangles equals the same scale factor kk that relates their corresponding sides — a fact used constantly to shortcut perimeter problems without finding every side first.

Step 1. Start from the definition. If △ABC∼△DEF\triangle ABC \sim \triangle DEF with scale factor kk, then by definition every pair of corresponding sides shares that ratio: ABDE=BCEF=CAFD=k\dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{CA}{FD} = k. Rearranging each equation gives AB=k⋅DEAB = k\cdot DE, BC=k⋅EFBC = k\cdot EF, CA=k⋅FDCA = k\cdot FD — each side of △ABC\triangle ABC is exactly kk times its partner in △DEF\triangle DEF.

Step 2. Write the perimeter of △ABC\triangle ABC using these substitutions. The perimeter is the sum of all three sides: PABC=AB+BC+CA=k⋅DE+k⋅EF+k⋅FDP_{ABC} = AB + BC + CA = k\cdot DE + k\cdot EF + k\cdot FD.

Step 3. Factor out kk, since it is common to all three terms: PABC=k (DE+EF+FD)P_{ABC} = k\,(DE + EF + FD). But DE+EF+FDDE + EF + FD is exactly PDEFP_{DEF}, the perimeter of △DEF\triangle DEF. So PABC=k⋅PDEFP_{ABC} = k \cdot P_{DEF}.

Step 4. Divide both sides by PDEFP_{DEF} to isolate the ratio: PABCPDEF=k\dfrac{P_{ABC}}{P_{DEF}} = k. The perimeters share the identical scale factor as the sides — no side length needs to be computed individually if only the perimeter ratio is wanted, and conversely a known perimeter ratio hands you kk for every side at once.

Worked example

In △ABC\triangle ABC and △PQR\triangle PQR, ∠A=∠P=50∘\angle A = \angle P = 50^\circ, ∠B=∠Q=70∘\angle B = \angle Q = 70^\circ, AB=8 cmAB = 8\text{ cm}, PQ=12 cmPQ = 12\text{ cm}, and BC=9 cmBC = 9\text{ cm}. Given △ABC∼△PQR\triangle ABC \sim \triangle PQR, find QRQR and ∠C\angle C.

Step 1. Find ∠C\angle C using the angle-sum property, since two angles of △ABC\triangle ABC are already known: ∠C=180∘−∠A−∠B=180∘−50∘−70∘=60∘\angle C = 180^\circ - \angle A - \angle B = 180^\circ - 50^\circ - 70^\circ = 60^\circ. (Consequently ∠R=60∘\angle R = 60^\circ too, by the similarity, though it wasn't asked.)

Step 2. Fix the correspondence before writing any ratio: A↔PA\leftrightarrow P, B↔QB\leftrightarrow Q, C↔RC\leftrightarrow R, so the side ratios that involve the known lengths are ABPQ=BCQR\dfrac{AB}{PQ}=\dfrac{BC}{QR}.

Step 3. Substitute the known values and solve for QRQR: 812=9QR  ⟹  QR=9×128=1088=13.5 cm\dfrac{8}{12}=\dfrac{9}{QR} \implies QR = \dfrac{9\times 12}{8} = \dfrac{108}{8} = 13.5\text{ cm}.

Answer: QR=13.5 cmQR = 13.5\text{ cm} and ∠C=60∘\angle C = 60^\circ.

Competitive stretch

△ABC∼△XYZ\triangle ABC \sim \triangle XYZ with scale factor k=35k = \dfrac{3}{5} (so △ABC\triangle ABC is the smaller triangle). The perimeter of △XYZ\triangle XYZ is 40 cm40\text{ cm} and XY=15 cmXY = 15\text{ cm}. Find the perimeter of △ABC\triangle ABC and the length of ABAB.

Step 1. Use the perimeter-ratio result just derived directly: Perimeter of △ABCPerimeter of △XYZ=k=35\dfrac{\text{Perimeter of }\triangle ABC}{\text{Perimeter of }\triangle XYZ} = k = \dfrac{3}{5}, so Perimeter of △ABC=35×40=24 cm\triangle ABC = \dfrac{3}{5}\times 40 = 24\text{ cm}.

Step 2. Find ABAB the same way, since ABAB corresponds to XYXY: ABXY=k=35  ⟹  AB=35×15=9 cm\dfrac{AB}{XY} = k = \dfrac{3}{5} \implies AB = \dfrac{3}{5}\times 15 = 9\text{ cm}.

Answer: perimeter of △ABC=24 cm\triangle ABC = 24\text{ cm} and AB=9 cmAB = 9\text{ cm} — both found without ever computing YZYZ or ZXZX.

Ask the tutor: If you're unsure which sides correspond in a similarity statement, ask the AI tutor to relabel your specific triangle pair and check your ratio setup before you compute.

Do similar triangles have to be the same size?

No — that would make them congruent, a stricter and different relationship. Similar triangles only need the same shape: equal corresponding angles and sides in one constant ratio, the scale factor kk. When k=1k=1 the two conditions happen to coincide and the triangles are both similar and congruent, but for any other kk the triangles are different sizes while still being similar. Congruence is really the special case of similarity where the scale factor is exactly 11.

Why does the order of letters in triangle ABC similar to triangle DEF matter so much?

Because that order is the only place the correspondence between vertices is recorded. Writing △ABC∼△DEF\triangle ABC \sim \triangle DEF tells you AA matches DD, BB matches EE, and CC matches FF — nothing else. If the true correspondence is actually A↔EA\leftrightarrow E instead, writing △ABC∼△DEF\triangle ABC \sim \triangle DEF would be simply false, and building a ratio like AB/DEAB/DE off the wrong pairing gives a wrong length even though every individual step of arithmetic is correct. Always check the correspondence against the given angle equalities before writing a single ratio.

If I only know that two pairs of angles are equal, do I need to check the sides too before calling the triangles similar?

No, and this is exactly what the AAA/AA similarity criterion (covered right after this topic) formalizes: for triangles specifically, matching angles alone is enough to guarantee the sides are automatically proportional, so you never need to measure sides to establish similarity from angles. The reverse also works — proportional sides alone (SSS) guarantee equal angles. That mutual shortcut is special to triangles and does not hold for general polygons like quadrilaterals, where equal angles do not force proportional sides.

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