IGCSE, Edexcel, IB and AP Science Courses | Dr Dabour Hub Academy

Overview

Define the similarity of two polygons and find the unknown side from the scale factor.
Prove two triangles similar with AA, SSS or SAS and use the result to find lengths.
Apply Euclid’s theorems and Pythagoras’ theorem with its converse in the right triangle.
Relate the ratio of areas to the square of the scale factor, and use the chord, secant and tangent theorems in the circle.

Similarity
├── 2-1 Ratio, proportion and similar polygons
│   ├── ratio and proportion (revision)
│   ├── definition of similarity (two conditions)
│   ├── scale factor k and unknown lengths
│   ├── polygons always similar / never certainly similar
│   └── map scale and models
├── 2-2 Similar triangles
│   ├── AA, SSS, SAS
│   ├── finding lengths and proving relations
│   ├── the right triangle and Euclid's two theorems
│   ├── Pythagoras and its converse
│   └── real-life applications (shadows, heights)
├── 2-3 Areas of similar figures
│   ├── ratio of areas = k²
│   ├── heights and perimeters
│   └── the line parallel to one side of a triangle
└── 2-4 Similarity in the circle
    ├── two intersecting chords
    ├── two secants from an external point
    ├── tangent and secant from an external point
    └── converses: four concyclic points, proving a tangent

Unit 2: Similarity of Polygons and Triangles, Areas and Circle Theorems · 1 / 12

2-1 Ratio and proportion (revision)

Comparison aspect	Main idea	Keywords
•	The ratio between two quantities of the same kind is the result of dividing one by the other written a : b or a/b b ≠ 0	ratiobetweenquantitiessame
A proportion is the equality of two ratios	a/b = c/d ⟺ ad = bc (cross-multiplication)	crossmultiplication
Ratio is an ordered relation, so a	b ≠ b : a unless a = b and it has no unit because the two quantities are of the same kind	unlessunitbecausequantities

Unit 2: Similarity of Polygons and Triangles, Areas and Circle Theorems · 2 / 12

2-2 Definition of the similarity of two polygons

Comparison aspect	Main idea	Example	Keywords
Two polygons are similar when both conditions hold	(1) corresponding angles are equal in measure and (2) corresponding sides are proportional	—	correspondinganglesequalmeasure
One condition alone is not enough	a rectangle and a square have equal angles but are not similar a rhombus and a square have proportional sides but are not similar	—	rectanglesquareequalangles
•	When the correspondence of vertices is written name the polygons in matching order	ABCD ~ PQRS means A↔P B↔Q …	correspondenceverticeswrittenname

Similarity of Polygons: two wooden frames with scale factor 2 show similarity of polygons: perimeters 14 and 28, areas 12 and 48.

Polygons that are always similar

Any two regular polygons with the same number of sides are similar, because a regular polygon is equilateral and equiangular.
Any two circles are similar. Two rectangles or two rhombi are not necessarily similar.

Unit 2: Similarity of Polygons and Triangles, Areas and Circle Theorems · 3 / 12

2-3 The scale factor, unknown lengths and map scale

The scale factor k is the ratio of a pair of corresponding sides; if k = 1 the two polygons are congruent.
To find an unknown length set the ratio of corresponding sides equal, then solve the proportion.

Comparison aspect	Main idea	Keywords
A scale 1	50000 means 1 cm on the map represents 50000 cm = 500 m in reality	50000meansrepresentsreality
•	Actual distance = map distance ÷ scale always convert units before computing	Actualdistancescalealways
Areas scale with the square of the scale	area on the map = actual area × (scale)²	actual

Unit 2: Similarity of Polygons and Triangles, Areas and Circle Theorems · 4 / 12

2-4 Cases of similarity of triangles

AA — two angles

If two angles of one triangle equal two angles of another, the triangles are similar. The third angle matches automatically because the angle sum of a triangle is 180°.

Order the sides of each triangle from smallest to largest
then compare the three ratios of corresponding sides
equality of all three gives similarity.

The equal angle must be the one included between the two proportional sides; otherwise the case does not apply.

Unit 2: Similarity of Polygons and Triangles, Areas and Circle Theorems · 5 / 12

2-5 Finding lengths and proving relations by similarity

Comparison aspect	Main idea	Example	Keywords
Step 1	look for equal angles (parallel lines vertically opposite angles angles on the same arc) or proportional sides	—	lookequalanglesparallel
Step 2	state the similarity with matching vertices in order	—	statesimilaritymatchingvertices
Step 3	write the ratio of corresponding sides and solve equal ratios give the unknown length and the product form gives relations	AB × DE = AC × DF	writeratiocorrespondingsides
•	Corresponding altitudes medians and angle bisectors of similar triangles are in the same ratio k as the sides	—	Correspondingaltitudesmediansangle

Unit 2: Similarity of Polygons and Triangles, Areas and Circle Theorems · 6 / 12

2-6 The right triangle and Euclid’s theorems

In right triangle ABC
right-angled at A
with altitude AD drawn to the hypotenuse BC
the two smaller triangles are similar to each other and to the whole triangle.

Leg theoremthe square of a leg equals the hypotenuse times the projection of that leg on the hypotenuse: AB² = BD × BC and AC² = CD × BC.

Altitude theoremthe square of the altitude to the hypotenuse equals the product of the two projections: AD² = BD × DC.

AB/AC = √(BD/DC); sin and cos of the acute angles read directly from the projections.
BD + DC = BC, so the two projections add to the hypotenuse.

Unit 2: Similarity of Polygons and Triangles, Areas and Circle Theorems · 7 / 12

2-7 Pythagoras’ theorem and its converse

The theorem

In a right triangle, the square on the hypotenuse equals the sum of the squares on the two legs.
Proof from Euclid AB² + AC² = BD·BC + CD·BC = (BD + CD)·BC = BC².

If the square of the longest side of a triangle equals the sum of the squares of the other two sides, then the triangle is right-angled, and the right angle lies opposite the longest side.
Compare instead of computing roots if c² > a² + b² the angle opposite c is obtuse; if c² < a² + b² it is acute.

Sun shadows light rays are parallel, so a person and his shadow form a triangle similar to that of a building and its shadow — heights are found from the ratio of shadows.
Ladders, diagonals of screens and fields, and the distance between two points on a grid.

Unit 2: Similarity of Polygons and Triangles, Areas and Circle Theorems · 8 / 12

2-8 Areas, heights and the line parallel to one side

The ratio of the areas of two similar polygons equals the square of the scale factor: A₁/A₂ = k².
The ratio of perimeters equals k itself; so from an area ratio you get k by taking the square root, and from k you get an unknown height or perimeter.

If DE ∥ BC in triangle ABC, then △ADE ~ △ABC, and the trapezoid DBCE has area = area(△ABC) − area(△ADE).
Hence area(△ADE)/area(DBCE) = AD² (AB² − AD²) — a standard exam form that asks for a ratio to the trapezoid, not to the whole triangle.

Unit 2: Similarity of Polygons and Triangles, Areas and Circle Theorems · 9 / 12

2-9 Similarity inside the circle

Two intersecting chords

If chords AB and CD meet at an interior point E, then △AEC ~ △DEB (vertically opposite angles at E and angles in the same segment standing on arc CB), and EA × EB = EC × ED.

From a point M outside the circle, two secants cut the circle at A, B and C, D where A and C are the nearer points and B and D the farther ones: MA × MB = MC × MD.

When the second secant collapses onto the tangent (C = D = T) the result becomes MT² = MA × MB — the square of the tangent equals the product of the segments of the secant.

Unit 2: Similarity of Polygons and Triangles, Areas and Circle Theorems · 10 / 12

2-10 The converses: four concyclic points and proving a tangent

If two points A and B lie on the same side of segment CD and subtend equal angles at it (∠CAD = ∠CBD), then the four points A, B, C, D lie on one circle.
If two chords AB and CD meet at E and EA × EB = EC × ED, then A, B, C, D are concyclic.
If from an external point M a line MT meets the circle at T and MT² = MA × MB for a secant MAB, then MT is a tangent to the circle at T.
The converse of the intersecting-chords product is the usual route in “prove that the four points lie on a circle” items.

Unit 2: Similarity of Polygons and Triangles, Areas and Circle Theorems · 11 / 12

Exam traps and key takeaways

💡 High-yield traps

Similarity needs both conditions for polygons; equal angles alone or proportional sides alone are not enough (square vs rectangle, square vs rhombus).
In SAS similarity the angle must be included between the two proportional sides.
Ratio of areas is k², ratio of lengths and perimeters is k — do not use one for the other.
The map scale is a ratio of lengths; areas need its square.
Write the similarity statement with corresponding vertices in the correct order, otherwise every later ratio is wrong.
Euclid’s theorems apply only after drawing the altitude to the hypotenuse of a right triangle.
In the secant theorem use the distance from the external point to each intersection, i.e. the whole secant times its external part.
MT² = MA·MB is the tangent–secant case; the intersecting-chords product uses two interior segments through the crossing point.
For “four points on one circle”, check the two angles subtend the same segment and lie on the same side of it.

Unit 2: Similarity of Polygons and Triangles, Areas and Circle Theorems · 12 / 12

Curriculum

Mathematics — First Secondary

Unit 1

Unit 2

Unit 3

Unit 4

Unit 2: Similarity of Polygons and Triangles, Areas and Circle Theorems