If one side of a triangle is extended, then the exterior angle is equal to the two opposite interior angles; and the three interior angles of a triangle are equal to two right angles. |
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Let ABC be a triangle, with side BC extended to D. Then the exterior angle ACD is equal to the two interior and opposite angles CAB, ABC; and the three interior angles ABC, BCA, CAB are together equal to two right angles. |
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Through the point C draw CE parallel to AB. |
(I. 31) |
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Then, since AB is parallel to CE, and AC meets them, |
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the alternate angles BAC, ACE are equal. |
(I. 29) |
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Also, since AB is parallel to CE and BD meets them, |
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angle ECD, which is exterior to those parallels, is equal to the opposite interior angle CBA; |
(I. 29) |
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and we proved that angle ACE was equal to angle BAC; |
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therefore the whole exterior angle ACD is equal to the two interior angles CBA, BAC. |
(Axiom 2) |
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To each of these equals add angle ACB; |
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therefore angles ACD, ACB are equal to the three angles CBA, BAC, ACB. |
(Axiom 2) |
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But angles ACD, ACB are equal to two right angles; |
(I. 13) |
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therefore the three angles CBA, BAC, ACB are together also equal to two right angles. |
(Axiom 1) |
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Therefore, if a side of a triangle etc. Q.E.D. |