Dem.—If both pairs of lines be coplanar, the proposition is the same as I. xxix., Ex. 2. If not, take any points A, C in the lines AB, BC, and cut off ED = BA, and EF = BC [I. iii.]. Join AD, BE, CF, AC, DF. Then because AB is equal and parallel to DE, AD is equal and parallel to BE [I. xxxiii]. In like manner CF is equal and parallel to BE. Hence [XI. ix.] AD is equal and parallel to CF. Hence [I. xxxiii.] AC is equal to DF. Therefore the triangles ABC, DEF, have the three sides of one respectively equal to the three sides of the other. Hence [I. viii.] the angle ABC is equal to DEF. Def. viii.—Two planes which meet are perpendicular to each other, when the right lines drawn in one of them perpendicular to their common section are normals to the other. Def. ix.—When two planes which meet are not perpendicular to each other, their inclination is the acute angle contained by two right lines drawn from any point of their common section at right angles to it—one in one plane, and the other in the other. Observation.—These definitions tacitly assume the result of Props. iii. and x. of this book. On this account we have departed from the usual custom of placing them at the beginning of the book. We have altered the place of Definition vi. for a similar reason. PROP. XI.—Problem. To draw a normal to a given plane (BH) from a given point (A) not in it. 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.