Kitap 1
15
Again, since the straight line DE stands on the straight line AB, making the angles AED, DEB, the angles AED, DEB are equal to two right angles. [I. 13]
15
But the angles CEA, AED were also proved equal to two right angles; therefore the angles CEA, AED are equal to the angles AED DEB. [Post. 4 and C. N. 1] Let the angle AED be subtracted from each; therefore the remaining angle CEA is equal to the remaining angle BED. [C. N. 3]
15
Similarly it can be proved that the angles CEB, DEA are also equal.
15
Therefore etc. Q. E. D.
15
[Porism. From this it is manifest that, if two straight lines cut one another, they will make the angles at the point of section equal to four right angles.
Proposition 16.
16
Enunciation In any triangle, if one of the sides be produced, the exterior angle is greater than either of the interior and opposite angles.
16
Proof. Let ABC be a triangle, and let one side of it BC be produced to D;
16
I say that the exterior angle ACD is greater than either of the interior and opposite angles CBA, BAC.
16
Let AC be bisected at E [I. 10], and let BE be joined and produced in a straight line to F;
16
let EF be made equal to BE[I. 3], let FC be joined [Post. 1], and let AC be drawn through to G [Post. 2].
16
Then, since AE is equal to EC, and BE to EF, the two sides AE, EB are equal to the two sides CE, EF respectively; and the angle AEB is equal to the angle FEC, for they are vertical angles. [I. 15] Therefore the base AB is equal to the base FC, and the triangle ABE is equal to the triangle CFE, and the remaining angles are equal to the remaining angles respectively, namely those which the equal sides subtend; [I. 4] therefore the angle BAE is equal to the angle ECF.
16
But the angle ECD is greater than the angle ECF; [C. N. 5] therefore the angle ACD is greater than the angle BAE.
16
Similarly also, if BC be bisected, the angle BCG, that is, the angle ACD [I. 15], can be proved greater than the angle ABC as well.
16
Therefore etc.
16
Q. E. D.
Proposition 17.
