Kitap 1
37
Moreover the triangle ABC is half of the parallelogram EBCA; for the diameter AB bisects it. [I. 34]
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And the triangle DBC is half of the parallelogram DBCF; for the diameter DC bisects it. [I. 34]
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[But the halves of equal things are equal to one another.]
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Therefore the triangle ABC is equal to the triangle DBC.
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Therefore etc.
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Q. E. D.
Proposition 38.
38
Enunciation Triangles which are on equal bases and in the same parallels are equal to one another.
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Proof. Let ABC, DEF be triangles on equal bases BC, EF and in the same parallels BF, AD; I say that the triangle ABC is equal to the triangle DEF.
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For let AD be produced in both directions to G, H; through B let BG be drawn parallel to CA, [I. 31] and through F let FH be drawn parallel to DE.
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Then each of the figures GBCA, DEFH is a parallelogram; and GBCA is equal to DEFH;
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for they are on equal bases BC, EF and in the same parallels BF, GH. [I. 36]
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Moreover the triangle ABC is half of the parallelogram GBCA; for the diameter AB bisects it. [I. 34]
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And the triangle FED is half of the parallelogram DEFH; for the diameter DF bisects it. [I. 34]
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[But the halves of equal things are equal to one another.]
38
Therefore the triangle ABC is equal to the triangle DEF.
38
Therefore etc.
38
Q. E. D.
Proposition 39.
39
Enunciation Equal triangles which are on the same base and on the same side are also in the same parallels.
