Unit 2: Similarity — Similar Polygons and Triangles, Their Areas and Applications in the Circle

• 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.
• A proportion is the equality of two ratios: a/b = c/d ⟺ ad = bc (cross-multiplication).
• 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.

• Two polygons are similar when both conditions hold: (1) corresponding angles are equal in measure, and (2) corresponding sides are proportional.
• 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.
• When the correspondence of vertices is written, name the polygons in matching order, e.g. ABCD ~ PQRS means A↔P, B↔Q …

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.

• 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.

Scale of maps and models

• A scale 1 : 50000 means 1 cm on the map represents 50000 cm = 500 m in reality.
• Actual distance = map distance ÷ scale; always convert units before computing.
• Areas scale with the square of the scale: area on the map = actual area × (scale)².

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°.

SSS — three proportional sides

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

SAS — two proportional sides and the included angle

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

• Step 1: look for equal angles (parallel lines, vertically opposite angles, angles on the same arc) or proportional sides.
• Step 2: state the similarity with matching vertices in order.
• Step 3: write the ratio of corresponding sides and solve; equal ratios give the unknown length, and the product form gives relations such as AB × DE = AC × DF.
• Corresponding altitudes, medians and angle bisectors of similar triangles are in the same ratio k as the sides.

• 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.

The two theorems

• Leg theorem: the square of a leg equals the hypotenuse times the projection of that leg on the hypotenuse: AB² = BD × BC and AC² = CD × BC.
• Altitude theorem: the square of the altitude to the hypotenuse equals the product of the two projections: AD² = BD × DC.

Useful deductions

• 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.

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.

The converse

• 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.

Everyday applications

• 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.

• 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.

A line parallel to one side of a triangle

• 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.

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.

Two secants from an external point

• 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.

Tangent and secant from an external point

• 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.

• 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.