Book 1
40
Let ABC, CDE be equal triangles on equal bases BC, CE and on the same side.
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I say that they are also in the same parallels.
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For let AD be joined; I say that AD is parallel to BE.
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For, if not, let AF be drawn through A parallel to BE [I. 31], and let FE be joined.
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Therefore the triangle ABC is equal to the triangle FCE; for they are on equal bases BC, CE and in the same parallels BE, AF. [I. 38]
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But the triangle ABC is equal to the triangle DCE; therefore the triangle DCE is also equal to the triangle FCE, [C.N. 1] the greater to the less: which is impossible. Therefore AF is not parallel to BE.
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Similarly we can prove that neither is any other straight line except AD; therefore AD is parallel to BE.
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Therefore etc. Q. E. D.]
Proposition 41.
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Enunciation If a parallelogram have the same base with a triangle and be in the same parallels, the parallelogram is double of the triangle.
41
Proof. For let the parallelogram ABCD have the same base BC with the triangle EBC, and let it be in the same parallels BC, AE;
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I say that the parallelogram ABCD is double of the triangle BEC.
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For let AC be joined.
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Then the triangle ABC is equal to the triangle EBC; for it is on the same base BC with it and in the same parallels BC, AE. [I. 37]
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But the parallelogram ABCD is double of the triangle ABC; for the diameter AC bisects it; [I. 34] so that the parallelogram ABCD is also double of the triangle EBC.
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Therefore etc.
41
Q. E. D.
Proposition 42.
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Enunciation To construct, in a given rectilineal angle, a parallelogram equal to a given triangle.
42
Proof. Let ABC be the given triangle, and D the given rectilineal angle; thus it is required to construct in the rectilineal angle D a parallelogram equal to the triangle ABC.
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Let BC be bisected at E, and let AE be joined; on the straight line EC, and at the point E on it, let the angle CEF be constructed equal to the angle D; [I. 23] through A let AG be drawn parallel to EC, and [I. 31] through C let CG be drawn parallel to EF.
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Then FECG is a parallelogram.
