Definition D1
△ABC and △A′B′C′ are similar under the correspondence A↔A′, B↔B′, C↔C′ if the corresponding angles are respectively equal and the ratios of corresponding sides are all equal.
Proof.
(1)A correspondence fixes three pairs of vertices.convention
(2)The first condition is ∠A = ∠A′, ∠B = ∠B′, ∠C = ∠C′.convention
(3)Side lengths are greater than 0, so ratios of sides are defined.A12
(4)The second condition is AB/A′B′ = BC/B′C′ = CA/C′A′. The common value is called the ratio of similarity and written k.(3), convention
(5)One condition alone does not make similarity. A similarity condition is a way of checking only part and obtaining both conditions.(2), (4), convention
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Lemma L1
Two congruent triangles are similar with ratio 1.
Proof.
(1)If △ABC ≅ △A′B′C′, the corresponding angles are respectively equal and the corresponding sides have equal lengths.assumption, A11
(2)Since AB = A′B′, AB/A′B′ = A′B′/A′B′ = 1, and the other two ratios are 1 for the same reason.(1), A12, A16
(3)Both conditions of D1 are satisfied, so the triangles are similar with ratio 1.(1), (2), D1
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Lemma L2
When D is between A and B, the line through D parallel to BC meets AC at a point E between A and C, and ∠DAE = ∠BAC, ∠ADE = ∠ABC, ∠AED = ∠ACB.
Proof.
(1)There is a line ℓ through D parallel to BC.A3
(2)ℓ cuts side AB of △ABC at D and does not meet BC, so it meets side AC at a point E. E is between A and C.(1), assumption, A4
(3)D lies on ray AB and E on ray AC, so ∠DAE and ∠BAC are the angle formed by the same two rays.(2), assumption, convention
(4)DE ∥ BC and line AB cuts both lines, so the corresponding angles are equal. ∠ADE = ∠ABC(1), (2), A1
(5)Line AC cuts both lines, so for the same reason ∠AED = ∠ACB(1), (2), A1
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Lemma L3
For the D and E of L2, AD/AB = AE/AC = DE/BC
Proof.
(1)DE is parallel to side BC and cuts the other two sides, so AD/AB = AE/ACL2, A2
(2)There is a line through E parallel to AB.A3
(3)That line cuts side CA at E and does not meet AB, so it meets side CB at a point F.(2), L2, A4
(4)EF is parallel to side AB and cuts the other two sides, so CE/CA = CF/CB(3), A2
(5)E is between A and C, and F between B and C, so AE + EC = AC, BF + FC = BCL2, (3), A6
(6)AE/AC + CE/CA = (AE + EC)/AC = AC/AC = 1(5), A15, A16
(7)In the same way, BF/BC + CF/CB = 1(5), A15, A16
(8)The two values in (4) are equal, so the remainders in (6)(7) are also equal. AE/AC = BF/BC(4), (6), (7), E
(9)DE ∥ BF and DB ∥ EF, so quadrilateral DBFE is a parallelogram, hence DE = BFL2, (2), (3), A5, convention
(10)Therefore AD/AB = AE/AC = DE/BC(1), (8), (9)
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Lemma L4
For the D, E of L2, if △ADE ≅ △A′B′C′ (A↔A′, D↔B′, E↔C′), then △ABC and △A′B′C′ are similar under A↔A′, B↔B′, C↔C′.
Proof.
(1)By the congruence, A′B′ = AD, A′C′ = AE, B′C′ = DE, and ∠A′ = ∠DAE, ∠B′ = ∠ADE, ∠C′ = ∠AED.assumption, A11
(2)Since ∠DAE = ∠BAC, ∠ADE = ∠ABC, ∠AED = ∠ACB, we have ∠A′ = ∠A, ∠B′ = ∠B, ∠C′ = ∠C(1), L2
(3)Since AD/AB = AE/AC = DE/BC, A′B′/AB = A′C′/AC = B′C′/BC(1), L3
(4)Both conditions of D1 are satisfied, so the two triangles are similar.(2), (3), D1
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Theorem T1
If ∠A = ∠A′ and ∠B = ∠B′, then △ABC and △A′B′C′ are similar.
Proof.
(1)Exchanging the names of the two triangles preserves the hypothesis, so we may assume A′B′ ≤ AB.assumption, convention
(2)Case A′B′ = AB: two angles and the included side are equal, so △ABC ≅ △A′B′C′, and by L1 they are similar.(1), assumption, A8, L1
(3)Case A′B′ < AB: take the point D on ray AB with AD = A′B′. D is between A and B.(1), A7, A6
(4)The line through D parallel to BC meets AC at E, and ∠DAE = ∠A, ∠ADE = ∠B(3), L2
(5)By the hypothesis, ∠DAE = ∠A′, ∠ADE = ∠B′, and the included side is AD = A′B′(3), (4), assumption
(6)Two angles and the included side are equal, so △ADE ≅ △A′B′C′(5), A8
(7)By L4, △ABC and △A′B′C′ are similar.(3), (4), (6), L4
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Theorem T2
If ∠A = ∠A′ and AB/A′B′ = AC/A′C′, then △ABC and △A′B′C′ are similar.
Proof.
(1)The proportion in the hypothesis is equivalent to A′B′/AB = A′C′/AC, so after renaming we may assume A′B′ ≤ AB.assumption, A13, A14, A12, convention
(2)Case A′B′ = AB: the hypothesis gives AB·A′C′ = AC·A′B′, so A′C′ = AC.(1), assumption, A13, A14, A12, E
(3)Two sides and the included angle are equal, so △ABC ≅ △A′B′C′, and by L1 they are similar.(2), assumption, A9, L1
(4)Case A′B′ < AB: take the point D on ray AB with AD = A′B′. D is between A and B.(1), A7, A6
(5)The line through D parallel to BC meets AC at E, and AD/AB = AE/AC(4), L2, L3
(6)AD = A′B′ and the hypothesis gives A′B′/AB = A′C′/AC, so AE/AC = A′C′/AC(4), (5), assumption, A13, A14, A12
(7)AE·AC = A′C′·AC and AC > 0, so AE = A′C′(6), A13, A12, E
(8)∠DAE = ∠A = ∠A′ and AD = A′B′, AE = A′C′, so △ADE ≅ △A′B′C′(4), (5), (7), assumption, L2, A9
(9)By L4, △ABC and △A′B′C′ are similar.(4), (5), (8), L4
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Theorem T3
If AB/A′B′ = BC/B′C′ = CA/C′A′, then △ABC and △A′B′C′ are similar.
Proof.
(1)The proportion in the hypothesis holds inverted as well, so after renaming we may assume A′B′ ≤ AB.assumption, A13, A14, A12, convention
(2)Case A′B′ = AB: the hypothesis gives B′C′ = BC and C′A′ = CA, so the three sides are respectively equal.(1), assumption, A13, A14, A12, E
(3)The three sides are equal, so △ABC ≅ △A′B′C′, and by L1 they are similar.(2), A10, L1
(4)Case A′B′ < AB: take the point D on ray AB with AD = A′B′. D is between A and B.(1), A7, A6
(5)The line through D parallel to BC meets AC at E, and AD/AB = AE/AC = DE/BC(4), L2, L3
(6)AD = A′B′, so the hypothesis and (5) give AE/AC = A′C′/AC, DE/BC = B′C′/BC(4), (5), assumption, A13, A14, A12
(7)AC > 0, BC > 0, so AE = A′C′, DE = B′C′(6), A13, A12, E
(8)The three sides are respectively equal, so △ADE ≅ △A′B′C′(4), (7), A10
(9)By L4, △ABC and △A′B′C′ are similar.(4), (5), (8), L4
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