Definition D1
For two distinct points A, B, the midpoint M of segment AB is the point that lies on segment AB and satisfies AM = MB.
Proof.
(1)Set A = (x₁, y₁), B = (x₂, y₂), with A ≠ B.convention
(2)The midpoint is defined by distance, not as the average of coordinates.convention
∎
Lemma L1
If 0 ≤ t ≤ 1 and P = A + t(B − A), then AP = t·AB and PB = (1 − t)·AB
Proof.
(1)P = (x₁ + t(x₂ − x₁), y₁ + t(y₂ − y₁)).A2
(2)The coordinate differences of A and P are t(x₂ − x₁) and t(y₂ − y₁).(1), A5, A6, A7
(3)AP = √(t²(x₂−x₁)² + t²(y₂−y₁)²)(2), A1, A8
(4) = √(t²((x₂−x₁)² + (y₂−y₁)²))(3), A4
(5) = t√((x₂−x₁)² + (y₂−y₁)²) = t·AB(4), A3, A1
(6)The coordinate difference of P and B is (x₂−x₁) − t(x₂−x₁) = (1 − t)(x₂−x₁), and likewise for y.(1), A4, A5, A6, A7, A9
(7)Since 0 ≤ 1 − t, the same method gives PB = (1 − t)·AB(6), A13, A1, A3, A4, A8
∎
(1)Write 1 + 1 as 2. Since 2 ≠ 0, there is a ½ with 2 × ½ = 1.convention, A14, A10
(2)½ + ½ = 1×½ + 1×½ = (1 + 1)½ = 2 × ½(1), A9, A4, A8
(3)Hence ½ + ½ = 1(1), (2)
∎
Lemma L3
A + ½(B − A) = ((x₁+x₂)/2, (y₁+y₂)/2)
Proof.
(1)x₁ = 1 × x₁ = (½ + ½)x₁ = ½x₁ + ½x₁L2, A9, A4, A8
(2)½(x₂ − x₁) = ½x₂ − ½x₁A4, A7, A8
(3)x₁ + ½(x₂ − x₁) = ½x₁ + ½x₂ = ½(x₁ + x₂)(1), (2), A4, A5, A6, A7
(4)Write ½(x₁ + x₂) as (x₁ + x₂)/2. The y-coordinate goes the same way.(3), convention
∎
Theorem T1
M₀ = ((x₁+x₂)/2, (y₁+y₂)/2) is the midpoint of segment AB.
Proof.
(1)Since 0 < ½ < 1, the point A + ½(B − A) lies on segment AB, and that point is M₀.A14, A2, L3
(2)AM₀ = ½·AB and M₀B = (1 − ½)·AB(1), L1
(3)1 − ½ = ½.L2, A6, A7
(4)Hence AM₀ = M₀B, and since M₀ lies on the segment, it is the midpoint.(1), (2), (3), D1
∎
Theorem T2
The midpoint of segment AB is unique.
Proof.
(1)Let M be a midpoint.assumption, D1
(2)M lies on segment AB, so there is a t with 0 ≤ t ≤ 1 and M = A + t(B − A).(1), D1, A2
(3)AM = t·AB and MB = (1 − t)·AB(2), L1
(4)Since AM = MB, we have t·AB = (1 − t)·AB(1), (3), D1
(5)Since A ≠ B, AB ≠ 0, and cancelling gives t = 1 − t(4), A12, A11, A8
(6)Adding t to both sides gives t + t = 1(5), E, A6, A7
(7)t + t = 1×t + 1×t = (1 + 1)t = 2t, so 2t = 1(6), A9, A4, A8
(8)Since 2 × ½ = 1 and 2 ≠ 0, we get t = ½(7), A10, A14, A11
(9)M = A + ½(B − A) = M₀(8), (2), L3
(10)The midpoint is M₀ alone, and its coordinates are the averages of the coordinates of the two points.(9), T1
∎