Let the two straight lines AB, CD cut one another at the point E; the angle AEC shall be equal to the angle DEB, and the angle CEB to the angle AED.
Because the straight line AE makes with the straight line CD the angles CEA, AED, those angles are together equal to two right angles. [I. 13.
Again, because the straight line DE makes with the straight line AB the angles AED, DEB, these also are together equal to two right angles. [I. 13.
But the angles CEA, AED have been shewn to be together equal to two right angles.
Therefore the angles CEA,AED are equal to the angles AED, DEB.
From each of these equals take away the common angle AED, and the remaining angle CEA is equal to the remaining angle DEB. [Axiom 3.
In the same manner it may be shewn that the angle CEB is equal to the angle AED.
Wherefore, if two straight lines &c. q.e.d.
Corollary 1. From this it is manifest that, if two straight lines cut one another, the angles which they make at the point where they cut, are together equal to four right angles.
Corollary 2. And consequently, that all the angles made by any number of straight lines meeting at one point, are together equal to four right angles.
PROPOSITION 16. THEOREM.
If one side of a triangle be produced, the exterior angle shall be greater than either of the interior opposite angles.
Let ABC be a triangle, and let one side BC be produced to D: the exterior angle ACD shall be greater than either of the interior opposite angles CBA, BAC.
Bisect AC at E, [I. 10.
join BE and produce it to F, making EF equal to EB, [I. 3.
and join FC.
Because AE is equal to EC, and BE to EF; [Constr.
the two sides AE, EB are equal to the two sides CE, EF, each to each;
