Definition D1
Two triangles with corresponding angles respectively equal and all ratios of corresponding sides equal to k are said to be similar with ratio k.
Proof.
(1)We use the definition of Proof 53 verbatim.A3, convention
(2)Side lengths are greater than 0, so k > 0.(1), A17
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Definition D2
A dilation of ratio k is a bijection f of the whole plane or space such that f(P)f(Q) = k·PQ for all points P, Q. Two figures F, F′ are similar with ratio k if F′ = f(F) for such an f.
Proof.
(1)f is a bijection, so it transfers inclusion and non-overlapping unchanged.convention
(2)The image of three points P, Q, R has all three corresponding side ratios equal to k, so it is similar to the original triangle, and hence the corresponding angles are equal.convention, A2, A3
(3)If PQ + QR = PR, the same relation holds in the image, so f preserves betweenness.convention, A14, A15
(4)Therefore the image of a square of side s is a square of side ks, and the image of a cube of edge s is a cube of edge ks.(2), (3), convention
(5)A triangle and its image are similar with ratio k in the sense of D1 as well. The two definitions do not conflict on triangles.(2), A2, A3, D1
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Lemma L1
In two triangles similar with ratio k, the ratio of the heights to corresponding bases is also k.
Proof.
(1)Suppose △ABC and △A′B′C′ are similar with ratio k. The corresponding angles are respectively equal and all corresponding side ratios equal k.assumption, D1, A3
(2)A triangle has at least two acute angles, so after renaming vertices we may take ∠B and ∠C acute. The corresponding angles are equal, so ∠B′, ∠C′ are also acute.(1), A13, convention
(3)The foot H of the perpendicular from A to line BC exists uniquely, and since ∠B and ∠C are acute, H is between B and C. The same holds from A′.(2), A11, A12
(4)△ABH and △A′B′H′ have right angles at H, H′, and H lies on ray BC, so ∠ABH = ∠B = ∠B′ = ∠A′B′H′.(1), (3), convention
(5)Two angles are respectively equal, so △ABH and △A′B′H′ are similar.(4), A1
(6)By similarity, AH/A′H′ = AB/A′B′ = k. AH and A′H′ are the heights to the corresponding bases BC, B′C′.(5), (1), (3), A3
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Theorem T1
If two triangles similar with ratio k have areas S, S′, then S/S′ = k²
Proof.
(1)As in L1, take bases BC, B′C′ and the heights AH, A′H′ to them.L1
(2)BC/B′C′ = k and AH/A′H′ = k, so BC = k·B′C′, AH = k·A′H′(1), D1, L1, A16, A17
(3)S = BC × AH ÷ 2, S′ = B′C′ × A′H′ ÷ 2(1), A4
(4)S = (k·B′C′)(k·A′H′) ÷ 2 = k²(B′C′ × A′H′ ÷ 2) = k²S′(2), (3), A15, E
(5)Since S′ > 0, S/S′ = k²(4), A17, A16
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Theorem T2
For two plane figures F, F′ similar with ratio k, the ratio of their areas is k².
Proof.
(1)Take a dilation f of ratio k with F′ = f(F), and let ε > 0 be given.assumption, D2
(2)There are unions of non-overlapping squares I, J with I ⊆ F ⊆ J and area(J) − area(I) ≤ ε(1), A9
(3)Writing sᵢ for the sides of the squares in I and tⱼ for those in J, area(I) = Σsᵢ², area(J) = Σtⱼ².(2), A5, A7
(4)f(I) and f(J) are unions of squares of sides ksᵢ, ktⱼ respectively, and f(I) ⊆ F′ ⊆ f(J)(2), (3), D2
(5)area(f(I)) = Σ(ksᵢ)² = k²Σsᵢ² = k²·area(I), and in the same way area(f(J)) = k²·area(J).(3), (4), A5, A7, A15
(6)area(I) ≤ area(F) ≤ area(J) and k²area(I) ≤ area(F′) ≤ k²area(J)(2), (4), (5), A8
(7)Both numbers lie between k²area(I) and k²area(J), so |area(F′) − k²area(F)| ≤ k²(area(J) − area(I)) ≤ k²ε(2), (6), A15, A18
(8)ε was arbitrary, so area(F′) = k²·area(F)(7), A10
(9)Since area(F) > 0, area(F′)/area(F) = k²(8), A17, A16
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Theorem T3
For two solids G, G′ similar with ratio k, the ratio of their volumes is k³.
Proof.
(1)The image of a cube of edge s is a cube of edge ks, and its volume is (ks)³ = k³s³.D2, A6, A15
(2)Replacing squares by cubes, area by volume, and k² by k³, lines (1)–(8) of T2 hold verbatim.(1), T2, A9, A7, A8, A10, A15, A18
(3)Since volume(G) > 0, volume(G′)/volume(G) = k³(2), A17, A16
(4)The square and the cube come not from the number of directions, but from the fact that when the side of a piece is scaled by k, the area of a square scales by k² and the volume of a cube by k³.(1), A5, A6
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