If at a point (B) in a right line (BA) two other right lines (CB, BD) on opposite sides make the adjacent angles (CBA, ABD) together equal to two right angles, these two right lines form one continuous line.
Dem.—If BD be not the continuation of CB, let BE be its continuation. Now, since CBE is a right line, and BA stands on it, the sum of the angles CBA, ABE is two right angles (xiii.); and the sum of the angles CBA, ABD is two right angles (hyp.); therefore the sum of the angles CBA, ABE is equal to the sum of the angles CBA, ABD. Reject the angle CBA, which is common, and we have the angle ABE equal to the angle ABD—that is, a part equal to the whole—which is absurd. Hence BD must be in the same right line with CB.
PROP. XV.—Theorem.
If two right lines (AB, CD) intersect one another, the opposite angles are equal (CEA = DEB, and BEC = AED).
Dem.—Because the line AE stands on CD, the sum of the angles CEA, AED is two right angles [xiii.]; and because the line CE stands on AB, the sum of the angles BEC, CEA is two right angles; therefore the sum of the angles CEA, AED is equal to the sum of the angles BEC, CEA. Reject the angle CEA, which is common, and we have the angle AED equal to BEC. In like manner, the angle CEA is equal to DEB.
The foregoing proof may be briefly given, by saying that opposite angles are equal because they have a common supplement.
Questions for Examination on Props. XIII., XIV., XV.
1. What problem is required in Euclid’s proof of Prop. xiii.?
2. What theorem? Ans. No theorem, only the axioms.
3. If two lines intersect, how many pairs of supplemental angles do they make?