PROP. I.—Theorem.
One part (AB) of a right line cannot be in a plane (X), and another part (BC) not in it.
Dem.—Since AB is in the plane X, it can be produced in it [Bk. I. Post. ii.]; let it be produced to D. Then, if BC be not in X, let any other plane passing through AD be turned round AD until it passes through the point C. Now, because the points B, C are in this second plane, the line BC [I., Def. vi.] is in it. Therefore the two right lines ABC, ABD lying in one plane have a common segment AB, which is impossible. Therefore, &c.
PROP. II.—Theorem.
Two right lines (AB, CD) which intersect one another in any point (E) are coplanar, and so also are any three right lines (EC, CB, BE) which form a triangle.
Dem.—Let any plane pass through EB, and be turned round it until it passes through C. Then because the points E, C are in this plane, the right line EC is in it [I., Def. vi.]. For the same reason the line BC is in it. Therefore the lines EC, CB, BE are coplanar; but AB and CD are two of these lines. Hence AB and CD are coplanar.
PROP. III.—Theorem.
If two planes (AB, BC) cut one another, their common section (BD) is a right line.
Dem.—If not from B to D, draw in the plane AB the right line BED, and in the plane BC the right line BFD. Then the right lines BED, BFD enclose a space, which [I., Axiom x.] is impossible. Therefore the common section BD of the two planes must be a right line.