Kitap 11
38
For the same reason, BSG is also a straight line, and BS is equal to SG.
38
Now, since CA is equal and parallel to DB, while CA is also equal and parallel to EG, therefore DB is also equal and parallel to EG. [XI. 9]
38
And the straight lines DE, BG join their extremities; therefore DE is parallel to BG. [I. 33]
38
Therefore the angle EDT is equal to the angle BGT, for they are alternate; [I. 29] and the angle DTU is equal to the angle GTS. [I. 15]
38
Therefore DTU, GTS are two triangles which have two angles equal to two angles, and one side equal to one side, namely that subtending one of the equal angles, that is, DU equal to GS, for they are the halves of DE, BG; therefore they will also have the remaining sides equal to the remaining sides. [I. 26]
38
Therefore DT is equal to TG, and UT to TS.
38
Therefore etc. Q. E. D.
PROPOSITION 39.
39
If there be two prisms of equal height, and one have a parallelogram as base and the other a triangle, and if the parallelogram be double of the triangle, the prisms will be equal.
39
Let ABCDEF, GHKLMN be two prisms of equal height, let one have the parallelogram AF as base, and the other the triangle GHK, and let the parallelogram AF be double of the triangle GHK; I say that the prism ABCDEF is equal to the prism GHKLMN.
39
For let the solids AO, GP be completed.
39
Since the parallelogram AF is double of the triangle GHK, while the parallelogram HK is also double of the triangle GHK, [I. 34] therefore the parallelogram AF is equal to the parallelogram HK.
39
But parallelepipedal solids which are on equal bases and of the same height are equal to one another; [XI. 31] therefore the solid AO is equal to the solid GP.
39
And the prism ABCDEF is half of the solid AO, and the prism GHKLMN is half of the solid GP; [XI. 28] therefore the prism ABCDEF is equal to the prism GHKLMN.
39
Therefore etc. Q. E. D.
