Kitap 1
39
Proof. Let ABC, DBC be equal triangles which are on the same base BC and on the same side of it; [I say that they are also in the same parallels.]
39
And [For] let AD be joined; I say that AD is parallel to BC.
39
For, if not, let AE be drawn through the point A parallel to the straight line BC, [I. 31] and let EC be joined.
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Therefore the triangle ABC is equal to the triangle EBC; for it is on the same base BC with it and in the same parallels. [I. 37]
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But ABC is equal to DBC; therefore DBC is also equal to EBC, [C.N. 1] the greater to the less: which is impossible.
39
Therefore AE is not parallel to BC.
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Similarly we can prove that neither is any other straight line except AD; therefore AD is parallel to BC.
39
Therefore etc.
39
Q. E. D.
[Proposition 40.
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Equal triangles which are on equal bases and on the same side are also in the same parallels.
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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.
40
For let AD be joined; I say that AD is parallel to BE.
40
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]
40
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.
