From photograph scale factors to measuring a mountain you can never climb, similarity is the mathematics of shapes that match without needing to match in size. This companion walks through the Basic Proportionality Theorem and the AA/SSS/SAS shortcuts for proving triangles similar, with tools, flashcards and a CBSE-style quiz to test yourself.
~AASSSSASAD/DB=AE/EC
🏔️6.1 Same shape, any size: what similarity really means
You already know congruent figures - same shape AND same size, like two identical cut-outs. This chapter is about a looser, more useful idea: figures with the same shape but not necessarily the same size. These are called similar figures.
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How do you measure a mountain you can't climb?. Nobody has ever wrapped a measuring tape around Mount Everest or stretched one to the Moon. Heights and distances like these are found through indirect measurement, and the mathematical idea that makes indirect measurement possible is exactly what this chapter builds toward: the similarity of triangles.
By the end of this chapter, you'll be able to solve real problems like: a girl walks away from a lamp-post, how long is her shadow after 4 seconds? A 6 m pole casts a 4 m shadow, how tall is a tower casting a 28 m shadow at the same moment? Both are solved with the same idea: similar triangles.
🖼️6.2 Similar figures: the photograph test
All circles with the same radius are congruent. All squares with the same side length are congruent. But what about circles of different radii, or squares of different sizes? They're clearly not congruent, yet something about them still matches.
All congruent figures are similar, but similar figures need not be congruent. Every circle is similar to every other circle. Every square is similar to every other square. Every equilateral triangle is similar to every other equilateral triangle, regardless of size, as long as the shape matches exactly.
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The photograph test. Think about the same photo of the Taj Mahal printed in stamp size, passport size, and postcard size, different sizes, but unmistakably the same shape. That's similarity. Now compare two same-sized photos of one person, aged 10 and aged 40, same size, but a completely different shape. Same size does NOT mean similar; same shape does.
stamp
passport
postcard
When a photographer enlarges a 35 mm negative into a 45 mm or 55 mm print, every line segment in the photo grows by the exact same ratio, 45:35 (or 55:35). That constant ratio is called the scale factor (or Representative Fraction), the same idea used to design world maps and building blueprints.
Definition: similar polygons. Two polygons with the same number of sides are similar if: (i) their corresponding angles are equal, AND (ii) their corresponding sides are in the same ratio (proportion).
💡Activity: shadows and similarity
Hang a bulb from the ceiling, place a cardboard quadrilateral ABCD between the bulb and a table below, and its shadow A'B'C'D' appears on the table. Because light travels in straight lines from a single point, every shadow-vertex lies exactly on the ray from the bulb through the matching cardboard vertex.
Measuring both shapes confirms it: every corresponding angle is equal, and every corresponding side is in the same ratio (AB/A'B' = BC/B'C' = CD/C'D' = DA/D'A'), exactly satisfying the similarity definition.
⚠️Neither condition alone is enough for polygons. A square and a non-square rectangle have all matching angles (90 degrees each), but their sides aren't in the same ratio. Not similar. A square and a rhombus (with slanted angles) can have all matching side lengths, but their angles don't match. Not similar. Both conditions together are required for general polygons like quadrilaterals.
📐6.3 The Basic Proportionality (Thales) Theorem
A triangle is just a polygon with 3 sides, so the same similarity rules apply. But triangles turn out to have a special extra structure worth its own theorem, one credited to the ancient Greek mathematician Thales.
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Thales (c. 640-546 BCE). Thales observed that in two equiangular triangles (triangles with all matching angles), the ratio of any two corresponding sides is always the same. This insight underlies what's now called the Basic Proportionality Theorem.
D on AB, E on AC, DE ∥ BC
Theorem 6.1 (Basic Proportionality Theorem). If a line is drawn parallel to one side of a triangle, intersecting the other two sides at distinct points, the other two sides are divided in the same ratio: AD/DB = AE/EC (where D is on AB, E is on AC, and DE is parallel to BC).
SET UPIn triangle ABC, DE is parallel to BC, with D on AB and E on AC. Join BE and CD, and draw DM perpendicular to AC and EN perpendicular to AB.
AREA FORMULASar(ADE) = half of AD times EN, ar(BDE) = half of DB times EN, ar(ADE) = half of AE times DM, ar(DEC) = half of EC times DM.
FORM RATIOSar(ADE)/ar(BDE) = AD/DB, and ar(ADE)/ar(DEC) = AE/EC.
KEY OBSERVATIONTriangle BDE and triangle DEC sit on the same base DE, between the same parallels DE and BC, so ar(BDE) = ar(DEC).
CONCLUDESince both ratios share equal denominators, their numerators' ratios must match too: AD/DB = AE/EC.
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Why the area trick works. Two triangles sharing the same base and squeezed between the same pair of parallel lines always have identical height, and therefore identical area, no measuring required, just geometry. That single fact is the engine driving the whole proof.
Theorem 6.2 (Converse of the BPT). 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 literally Theorem 6.1 run in reverse.)
Worked example: proving a ratio through a trapezium
ABCD is a trapezium with AB parallel to DC. E and F sit on the non-parallel sides AD and BC, with EF parallel to AB. Show that AE/ED = BF/FC.
KEY MOVEJoin diagonal AC, meeting EF at G. Since EF is parallel to AB and AB is parallel to DC, it follows that EF is parallel to DC too.
APPLY BPT TWICEIn triangle ADC, EG is parallel to DC, so AE/ED = AG/GC. In triangle CAB, GF is parallel to AB, so CG/AG = CF/BF, i.e. AG/GC = BF/FC.
COMBINEBoth expressions equal AG/GC, so AE/ED = BF/FC.
Worked example: using the converse to prove a triangle is isosceles
Given PS/SQ = PT/TR and angle PST = angle PRQ, prove triangle PQR is isosceles.
APPLY CONVERSEPS/SQ = PT/TR means, by Theorem 6.2, that ST is parallel to QR.
CORRESPONDING ANGLESSince ST is parallel to QR, angle PST = angle PQR (corresponding angles).
COMBINE WITH GIVENWe're also given angle PST = angle PRQ. So angle PRQ = angle PQR.
CONCLUDEEqual base angles mean the sides opposite them are equal: PQ = PR. So triangle PQR is isosceles.
🔺6.4 Three shortcuts for proving triangles similar
Checking all three angles AND all three side ratios every single time would be exhausting. Just like congruence has shortcut criteria (SSS, SAS, ASA), similarity has its own shortcuts, and for triangles specifically, checking just one of the two conditions turns out to guarantee the other automatically.
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Like verifying a blueprint at a glance. You don't need to re-measure every wall of a building to confirm two blueprints represent the same design at different scales, a few key matching measurements (the right angles, or a couple of proportional wall lengths) are enough to be confident the whole shape matches. That's exactly what these three criteria give you for triangles.
Theorem 6.3, AAA (Angle-Angle-Angle) criterion. If corresponding angles of two triangles are equal, their corresponding sides are automatically in the same ratio, so the triangles are similar.
✂️You only need AA, not AAA. By the angle sum property, if two pairs of angles match, the third pair is forced to match too (all triangle angles sum to 180 degrees). So checking just two angle pairs, the AA criterion, is already enough.
Theorem 6.4, SSS (Side-Side-Side) criterion. If corresponding sides of two triangles are all in the same ratio, their corresponding angles are automatically equal, so the triangles are similar.
Theorem 6.5, SAS (Side-Angle-Side) criterion. If one angle of a triangle equals one angle of another, and the sides forming that angle are in the same ratio, the two triangles are similar.
Notation matters: similarity must respect vertex correspondence. If triangle ABC is similar to triangle DEF, that specifically means A matches D, B matches E, C matches F. Writing it in a different vertex order would claim a different (and possibly false) correspondence.
Worked example: vertical angles and parallel lines (AA)
If PQ is parallel to RS in a figure where PS and QR cross at O, prove triangle POQ is similar to triangle SOR.
ALTERNATE ANGLESPQ parallel to RS gives angle P = angle S and angle Q = angle R (alternate angles).
VERTICALLY OPPOSITEAngle POQ = angle SOR (vertically opposite angles), a bonus third confirmation.
CONCLUDETwo (in fact three) pairs of equal angles gives triangle POQ similar to triangle SOR by AA.
🏮The payoff: the lamp-post shadow problem
A 90 cm-tall girl walks away from a 3.6 m lamp-post at 1.2 m/s. Find her shadow's length after 4 seconds.
SET UPAfter 4 s, she's walked BD = 1.2 times 4 = 4.8 m from the post. Let her shadow length be x = DE.
ESTABLISH SIMILARITY (AA)In triangle ABE and triangle CDE: angle B = angle D = 90 degrees (post and girl both stand vertical to the ground), and angle E = angle E (shared angle). So triangle ABE is similar to triangle CDE by AA, no side measurements needed to establish the similarity.
USE THE PROPORTIONBE/DE = AB/CD, so (4.8+x)/x = 3.6/0.9 (using 90 cm = 0.9 m).
SOLVE4.8+x = 4x, so 3x = 4.8, so x = 1.6. Her shadow is 1.6 m long.
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The same trick, at mountain scale. A 6 m vertical pole casts a 4 m shadow; at the same moment, a tower casts a 28 m shadow. Since both shadows are cast by the same sun at the same angle, the pole-and-shadow triangle is similar to the tower-and-shadow triangle: height/shadow stays constant. Height of tower = 6 times (28/4) = 42 m, solved without ever touching the tower.
📎A useful extra: the RHS criterion. For right triangles specifically, there's a shortcut even simpler than SAS: if the hypotenuse and one other side of one right triangle are proportional to the hypotenuse and corresponding side of another right triangle, the two triangles are similar. This works because in a right triangle, the third side is never independent, it's always fixed by the Pythagorean relationship between the other two.
📋6.5 Summary
Similar figures have the same shape but not necessarily the same size. All congruent figures are similar, but similar figures need not be congruent.
Two polygons of the same number of sides are similar if corresponding angles are equal and corresponding sides are in the same ratio.
Theorem 6.1 (BPT): a line parallel to one side of a triangle divides the other two sides in the same ratio.
Theorem 6.2 (converse): a line dividing two sides of a triangle in the same ratio is parallel to the third side.
Three shortcuts prove triangle similarity without checking all six parts: AA/AAA (equal angles), SSS (proportional sides), SAS (one equal angle plus proportional including sides).
RHS criterion (a bonus, for right triangles): proportional hypotenuse plus one other side is enough.
TOOL 1 Triangle Similarity Checker
Enter two triangles' known angles and/or side lengths. The tool checks whether AA, SSS, or SAS conditions are satisfied, reports whether the triangles are similar, and names the criterion used.
✅ Similar by AA
Angles at positions 1 and 2 and 3 match — AA criterion satisfied.
Side ratios (T2/T1): 2, 2, 2
TOOL 2 Indirect Height / Shadow Calculator
Enter a reference object's height and shadow length, plus a second shadow length measured at the same moment, to compute the unknown height using similar triangles, the same method used to measure mountains and towers.
SET UP RATIO
height/shadow is constant: 6/4 = target height / 28
SOLVE
target height = 6 × (28 / 4) = 6 × 7 = 42
Target height: 42 (ratio 7)
§3 Flashcards
Click a card to flip it. Use Prev/Next to move through the deck.
Card 1 / 12
Q
What are similar figures?
A
Figures that have the same shape but not necessarily the same size.
§4 Quiz
Modelled on the CBSE Class 10 Section A paper — Multiple Choice and Assertion–Reason questions, each worth 1 mark. 20 questions per attempt, drawn from a pool of 40.
