Definition D1
Fix two transformations of the coordinate plane. For real numbers a, b, call T(x, y) = (x + a, y + b) a translation, and call σ(x, y) = (x, −y) the reflection in the x-axis.
Proof.
(1)Both transformations are determined at each point by its coordinates alone.convention
(2)T adds constants to both coordinates, and σ changes only the sign of the y-coordinate.(1), convention
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Definition D2
A transformation f of the plane is called an isometry if d(f(P), f(Q)) = d(P, Q) for every pair of points P, Q.
Proof.
(1)All this definition demands is the single distance between two points.convention
(2)Neither angles nor betweenness are in the definition. Whether they follow is the question of this walk.(1), convention
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Lemma L1
The translation T is an isometry.
Proof.
(1)Let P = (x₁, y₁), Q = (x₂, y₂). Then T(P) = (x₁+a, y₁+b), T(Q) = (x₂+a, y₂+b)D1
(2)(x₂+a) − (x₁+a) = (x₂ + a) + ((−x₁) + (−a)) = (x₂ + (−x₁)) + (a + (−a)) = x₂ − x₁A11, A12, A8, A9, A10
(3)In the same way, (y₂+b) − (y₁+b) = y₂ − y₁A11, A12, A8, A9, A10
(4)d(T(P), T(Q))² = ((x₂+a)−(x₁+a))² + ((y₂+b)−(y₁+b))²(1), A1
(5)= (x₂−x₁)² + (y₂−y₁)² = d(P, Q)²(2), (3), (4), A1, E
(6)The two distances are nonnegative with equal squares, so d(T(P), T(Q)) = d(P, Q)(5), A7
(7)This holds for every pair of points, so T is an isometry.(6), D2
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Lemma L2
The reflection σ in the x-axis is an isometry.
Proof.
(1)σ(P) = (x₁, −y₁), σ(Q) = (x₂, −y₂)D1
(2)The x-coordinates are unchanged, so their difference is x₂ − x₁(1)
(3)(−y₂) − (−y₁) = (−y₂) + y₁ = −(y₂ + (−y₁)) = −(y₂ − y₁)A11, A12, A9
(4)(−(y₂−y₁))² = (y₂−y₁)²A13
(5)d(σ(P), σ(Q))² = (x₂−x₁)² + (−(y₂−y₁))² = (x₂−x₁)² + (y₂−y₁)² = d(P, Q)²(1), (2), (3), (4), A1, E
(6)d(σ(P), σ(Q)) = d(P, Q)(5), A7
(7)This holds for every pair of points, so σ is an isometry.(6), D2
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Theorem T1
An isometry preserves angles.
Proof.
(1)Let f be an isometry and ∠BAC an angle. B and C are points other than A on the two sides of the angle.assumption, D2, A5
(2)f carries the three distances unchanged. AB = f(A)f(B), BC = f(B)f(C), CA = f(C)f(A)(1), D2
(3)Consider first the case where A, B, C are not collinear.assumption
(4)If f(A), f(B), f(C) were collinear, one would lie between the other two, making the sum of two distances equal the third. By (2) the same equation would hold for A, B, C, contradicting (3).(2), (3), A4, A6
(5)The three sides are respectively equal, so △ABC ≅ △f(A)f(B)f(C)(2), (4), A2
(6)Congruent, so the corresponding angles are equal. ∠BAC = ∠f(B)f(A)f(C)(5), A3
(7)If A, B, C are collinear, ∠BAC is 0° or 180°, and which one is determined by whether A lies between B and C.(1), A6, convention
(8)By (2) and the betweenness test, f carries that betweenness unchanged.(2), A4
(9)Hence the image angle is likewise 0° or 180°, the same one.(7), (8)
(10)An angle is determined by two rays, and the result is the same wherever B and C are taken, so (6) and (9) are properties of the angle itself.(6), (9), A5
(11)This block cited neither coordinates nor A1. The preservation of angles follows from SSS congruence alone.(5), (6), A2, A3
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Corollary C
The translation T and the reflection σ in the x-axis preserve both distances between points and angles.
Proof.
(1)T and σ are isometries.L1, L2
(2)An isometry preserves angles.T1
(3)Hence T and σ preserve both lengths and angles.(1), (2)
(4)The preservation of length came from coordinate computation and the distance formula; the preservation of angles came from SSS congruence. The two grounds are not the same.(1), (2), A1, A2
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