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7 Cards in this Set
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Lemma 4 'Side-angle-side': If ABC, A'B'C' are triangles s.t. |AB|=|A'B'|, |BC|=|B'C'|, β=β', then ABC and A'B'C' are congruent. |
|BC|=|B'C'| so ∃ an isometry f:E²→E² s.t. f(B)=B' f(C)=C' by axiom 6. By axiom 7, wlog f(A) lies on the same side of B'C' as A' does. Now, isometries preserve angles so f(β)=β=β', so f(A) lies on the line A'B'. Since isometries preserve distances, |f(A)f(B)|=|AB|=|A'B'|, so f(A)=A'. Therefore, ABC is congruent to A'B'C'.
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Lemma 5: Let L be a line in E². The reflection in L is an isometry. |
Picture required
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Lemma 6 'Angle-side-angle' : If ABC, A'B'C' are triangles s.t. |BC|=|B'C'|, β=β', γ=γ', then ABC and A'B'C' are congruent. |
|BC|=|B'C'|∴∃f isometry s.t. f(B)=B', f(C)=C' by axiom 6. By axiom 7, wlog f(A) is on the same side of the line B'C' as A' is. Isometries preserve angles, so ∠f(A)f(B)f(C) = ∠ABC=β=β' and ∠f(B)f(C)f(A) = ∠BCA=γ=γ' The line at angle β from B' meets the line at angle γ from C' at one point; A'. Since f(A) lies on both these lines, f(A)=A'. |
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Lemma 7 'Isosceles' : Let ABC be a triangle. Then |AB|=|BC| iff α=γ. |
Use ASA and SAS. |
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Lemma 8 'SSS' |
... ...(but this time on opposite side) Triangles A'f(A)C' and A'f(A)B' are isosceles. ... By SAS |
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Lemma 9 'Perpendicular Bisectors': Given A≠B, the perpendicular bisector of AB is the locus of points equidistant from A and B. |
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Theorem 10:If ABC is any triangle, then the perpendicular bisectors of the three edges are concurrent.Moreover, if O is that meeting point, then O is the centre of the circle going through A, B and C.
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