A line drawn parallel to one side of a triangle cuts the other two sides in the same ratio.
Precisely: If a line parallel to side BC of △ABC meets AB at D and AC at E, then AD/DB = AE/EC; conversely, a line dividing two sides in the same ratio is parallel to the third side.
How it works
Triangles ADE and ABC are equiangular (the parallel line makes equal corresponding angles), so their sides are in proportion. It is named after Thales (about 640 to 546 BCE), who is believed to have used it.
Examples
By the Basic Proportionality Theorem, if DE ∥ BC with AD = 1.5 cm, DB = 3 cm and AE = 1 cm, then EC = 2 cm.
The converse of the Basic Proportionality Theorem: if AD/DB = AE/EC, then DE ∥ BC.
Parallel lines cutting two straight roads divide them in the same ratio, by the Basic Proportionality Theorem.
Do not confuse: The ratio is of the parts of the sides (AD/DB), not of a part to the whole (AD/AB), although AD/AB = AE/AC is also true.
Used in
similar triangles, dividing a line segment in a given ratio (constructions), proofs in geometry
The product of the binomials x + 5 and x + 6 is x² + 11x + 30.
Do not confuse: A monomial has one term (7x), a binomial two (7x + 1), a trinomial three (x² + 7x + 1). Like terms must be added first: x + 2x is one term, 3x.
Used in
identities, factorisation, the binomial theorem (Class 11)
The rule for multiplying out (a + b)ⁿ: the terms are ⁿCᵣ aⁿ⁻ʳ bʳ for r = 0, 1, 2, …, n, with the powers of a falling as the powers of b rise.
Precisely: For a positive whole number n, (a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + … + ⁿCₙbⁿ; the (r + 1)-th term is ⁿCᵣ aⁿ⁻ʳ bʳ.
How it works
Multiplying n brackets (a + b), each term picks a or b from every bracket. The terms with b picked r times number ⁿCᵣ, and each is aⁿ⁻ʳbʳ, so that term has coefficient ⁿCᵣ.
Examples
By the binomial theorem, (a + b)³ = a³ + 3a²b + 3ab² + b³.
OpenStax: College Algebra 2e, 9.6 Binomial Theorem
Brahmagupta's formula
BRAH-muh-GOOP-tuhz FOR-myoo-luhClass 9
A way to find the area of a four-sided shape whose corners lie on a circle, from its four sides alone.
Precisely: For a cyclic quadrilateral with sides a, b, c and d and semi-perimeter s = (a + b + c + d)/2, area = √((s − a)(s − b)(s − c)(s − d)).
How it works
Found by Brahmagupta in 628 CE. It works only for cyclic 4-gons: the sides of a general 4-gon do not fix its area. Put d = 0 and the 4-gon becomes a triangle (every triangle is cyclic), and the formula turns into Heron's.
Examples
For a rectangle a by b, Brahmagupta's formula gives s = a + b and area √(b × a × b × a) = ab.
A trapezium with sides 40, 26, 20 and 26 cm is cyclic, and Brahmagupta's formula gives √(16 × 30 × 36 × 30) = 720 cm².
A cyclic 4-gon with sides 5, 5, 12 and 12 has s = 17, so Brahmagupta's formula gives √(12 × 12 × 5 × 5) = 60.
With d = 0, Brahmagupta's formula becomes Heron's formula for a triangle.
Do not confuse: It is only for cyclic 4-gons. A rhombus of side 3 that is not a square is not cyclic, and its area is not √(3⁴) = 9.
Used in
areas of cyclic quadrilaterals, a famous example of generalisation