Kitap 11
20
And, since the two sides BD, DC are greater than BC, [I. 20] and of these DB was proved equal to BE, therefore the remainder DC is greater than the remainder EC.
20
Now, since DA is equal to AE, and AC is common, and the base DC is greater than the base EC, therefore the angle DAC is greater than the angle EAC. [I. 25]
20
But the angle DAB was also proved equal to the angle BAE; therefore the angles DAB, DAC are greater than the angle BAC.
20
Similarly we can prove that the remaining angles also, taken together two and two, are greater than the remaining one.
20
Therefore etc. Q. E. D.
PROPOSITION 21.
21
Any solid angle is contained by plane angles less than four right angles.
21
Let the angle at A be a solid angle contained by the plane angles BAC, CAD, DAB; I say that the angles BAC, CAD, DAB are less than four right angles.
21
For let points B, C, D be taken at random on the straight lines AB, AC, AD respectively, and let BC, CD, DB be joined.
21
Now, since the solid angle at B is contained by the three plane angles CBA, ABD, CBD, any two are greater than the remaining one; [XI. 20] therefore the angles CBA, ABD are greater than the angle CBD.
21
For the same reason the angles BCA, ACD are also greater than the angle BCD, and the angles CDA, ADB are greater than the angle CDB; therefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are greater than the three angles CBD, BCD, CDB.
21
But the three angles CBD, BDC, BCD are equal to two right angles; [I. 32] therefore the six angles CBA, ABD, BCA, ACD, CDA, ADB are greater than two right angles.
21
And, since the three angles of each of the triangles ABC, ACD, ADB are equal to two right angles, therefore the nine angles of the three triangles, the angles CBA, ACB, BAC, ACD, CDA, CAD, ADB, DBA, BAD are equal to six right angles; and of them the six angles ABC, BCA, ACD, CDA, ADB, DBA are greater than two right angles; therefore the remaining three angles BAC, CAD, DAB containing the solid angle are less than four right angles.
21
Therefore etc. Q. E. D.
