If three concurrent lines (BC, BD, BE) have a common perpendicular (AB), they are coplanar.
Dem.—For if possible let BC be not coplanar with BD, BE, and let the plane of AB, BC intersect the plane of BD, BE in the line BF. Then [XI. iii.] BF is a right line; and, since it is coplanar with BD, BE, which are each perpendicular to AB, it is [XI. iv.] perpendicular to AB. Therefore the angle ABF is right; and the angle ABC is right (hyp.). Hence ABC is equal to ABF, which is impossible [I., Axiom ix.]. Therefore the lines BC, BD, BE are coplanar.
PROP. VI.—Theorem.
If two right lines (AB, CD) be normals to the same plane (X), they shall be parallel to one another.
Dem.—Let AB, CD meet the plane X at the points B, D. Join BD, and in the plane X draw DE at right angles to BD; take any point E in DE. Join BE, AE, AD. Then because AB is normal to X, the angle ABE is right. Therefore AE2 = AB2 + BE2 = AB2 + BD2 + DE2; because the angle BDE is right. But AB2 + BD2 = AD2, because the angle ABD is right. Hence AE2 = AD2 + DE2. Therefore the angle ADE is right. [I. xlviii]. And since CD is normal to the plane X, DE is perpendicular to CD. Hence DE is a common perpendicular to the three concurrent lines CD, AD, BD. Therefore these lines are coplanar [XI. v.]. But AB is coplanar with AD, BD [XI. ii.]. Therefore the lines AD, BD, CD are coplanar; and since the angles ABD, BDC are right, the line AB is parallel to CD [I. xxviii.].
Def. vii.—If from every point in a given line normals be drawn to a given plane, the locus of their feet is called the projection of the given line on the plane.
Exercises.
1. The projection of any line on a plane is a right line.
2. The projection on either of two intersecting planes of a normal to the other plane is perpendicular to the line of intersection of the planes.