For let any straight line AB set up on the straight line CD make the angles CBA, ABD;
I say that the angles CBA, ABD are either two right angles or equal to two right angles.

Now, if the angle CBA is equal to the angle ABD, they are two right angles. [Def. 10]

But, if not, let BE be drawn from the point B at right angles to CD; [I. 11]
therefore the angles CBE, EBD are two right angles.

Then, since the angle CBE is equal to the two angles CBA, ABE,
let the angle EBD be added to each;
therefore the angles CBE, EBD are equal to the three angles CBA, ABE, EBD. [C.N. 2]

Again, since the angle DBA is equal to the two angles DBE, EBA,
let the angle ABC be added to each;
therefore the angles DBA, ABC are equal to the three angles DBE, EBA, ABC. [C.N. 2]

But the angles CBE, EBD were also proved equal to the same three angles;
and things which are equal to the same thing are also equal to one another; [C.N. 1]
therefore the angles CBE, EBD are also equal to the angles DBA, ABC.
But the angles CBE, EBD are two right angles;
therefore the angles DBA, ABC are also equal to two right angles.

Therefore etc. Q.E.D.