The First Six Books of the Elements of Euclid

by Euclid; John Casey

Sol.—In the given plane BH draw any line BC, and from A draw AD perpendicular to BC [I. xii.]; then if AD be perpendicular to the plane, the thing required is done. If not, from D draw DE in the plane BH at right angles to BC [I. xi.], and from A draw AF [I. xii.] perpendicular to DE. AF is normal to the plane BH.

Dem.—Draw GH parallel to BC. Then because BC is perpendicular both to ED and DA, it is normal to the plane of ED, DA [XI. iv.]; and since GH is parallel to BC, it is normal to the same plane [XI. viii.]. Hence AF is perpendicular to GH [XI. Def. vi.], and AF is perpendicular to DE (const.). Therefore AF is normal to the plane of GH and ED—that is, to the plane BH.

PROP. XII.—Problem.

To draw a normal to a given plane from a given point (A) in the plane.

Sol.—From any point B not in the plane draw [XI. xi.] BC normal to it. If this line pass through A it is the normal required. If not, from A draw AD parallel to BC [I. xxxi.]. Then because AD and BC are parallel, and BC is normal to the plane, AD is also normal to it [XI. viii.], and it is drawn from the given point. Hence it is the required normal.

PROP. XIII.—Theorem.

From the same point (A) there can be but one normal drawn to a given plane (X).

Dem.—1. Let A be in the given plane, and if possible let AB, AC be both normals to it, on the same side. Now let the plane of BA, AC cut the given plane X in the line DE. Then because BA is a normal, the angle BAE is right. In like manner CAE is right. Hence BAE = CAE, which is impossible.