If one side of a triangle is extended, then the exterior
angle is greater than either of the opposite interior angles. |
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Let ABC be a triangle, and let one side of it BC be extended to D;
then the exterior angle ACD is greater than either of the
opposite interior angles, ABC, CAB. |
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(We now draw triangle BFC in such a way that triangles BAE, ECF will be congruent; this makes angle BAE equal to angle ECF. Eventually, Proposition 27, that will imply that the lines BA, CF are parallel.) |
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Bisect AC at E; |
(I. 10) |
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draw BE and extend it in a straight line to F, making EF equal to BE; |
(I. 3) |
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draw FC, and extend AC to a point G. |
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Then, since AE is equal to EC, and BE to EF, |
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the two sides AE, EB are equal to the two sides CE, EF respectively; |
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and angle AEB is equal to angle CEF,
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because they are vertical angles. |
(I. 15) |
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Therefore triangles AEB, CEF are congruent, |
(S.A.S.) |
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and the remaining angles are equal to the remaining angles, namely those that are opposite the equal sides: |
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angle BAE is equal to angle ECF. |
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But angle ECD is greater than angle ECF; |
(Axiom 5) |
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therefore angle ACD is greater than angle BAE. |
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If we bisect BC, in the same way then we can prove that angle BCG, |
that is, angle ACD, |
(I. 15) |
is greater than angle ABC as well. |
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Therefore, if one side of a triangle etc. Q.E.D. |