GEOMETRY. 



(3.) Let AB and CD be two parallel lines. By 

 the definition of parallel 

 lines, they are in the same 

 plane P. If we take upon 

 the line AB any two points 

 A and B, and in the line 

 CD any point C, the plane 

 P containing these three 



points not in a straight line is completely deter- 

 mined. 



THE LINE AND PLANE PERPENDICULAR TO 

 EACH OTHER. 



Definition, A straight line and a plane which 

 intersect are perpendicular to each other, when 

 the straight line is perpendicular to every straight 

 line which can be drawn through its foot in the 

 plane. 



PROP. II. A line which is perpendicular to 

 each of two intersecting lines at their point of 

 intersection, is perpendicular to every straight line 

 through the point in the plane which contains 

 them. 



Let AB be perpendicular to BD, and also per- 

 pendicular to BC at their point of intersection B ; 

 then it shall be perpendicular to any line BE 

 drawn through B in the 

 plane of BD and BC 

 Draw a line DEC, cut- 

 ting BD, BE, and BC in 

 D, E, and C, and join 

 each of these points with 

 A and A', two points in 

 AB at equal distances 

 from B. Then it is evi- 

 dent (Prop. VII. I, p. 



612) that AC = A'C, and AD = A'D. Com- 

 paring the two triangles ACD, A'CD, we have 

 three sides in the one equal to three sides in 

 the other, and therefore angle ACE == angle 

 A'CE. Again, comparing the two triangles 

 ACE, A'CE, we have two sides AC, CE in the 

 one equal to A'C, CE in the other, and the con- 

 tained angles ACE, A'CE equal, therefore AE 

 = A'E. Then in the two triangles ABE, A'BE, 

 we have the three sides of the one equal to the 

 three sides of the other, therefore the angle ABE 

 is equal to the angle A'BE ; therefore each is a 

 right angle, therefore AB is perpendicular to BE. 

 Hence a line is perpendicular to a plane, when 

 it is perpendicular to each of two lines drawn 

 through its foot in that plane. 



PROP. III. Through a given point to draw a 

 plane perpendicular to a given line. 



(i.) If the given point be on the line. 



Let AB be the given line, and O the given point 

 in it ; it is required to draw through O a plane 

 perpendicular to AB. From 

 the point O draw the two 

 lines OC and OD each per- 

 pendicular to AB, but in 

 different planes namely, 

 AOC and AOD ; then it is 

 evident that the plane of 

 OC and OD is the plane 

 required. 



given 



(2.) If the given point be without the given 

 line. 



From the point O, in the plane which is deter- 

 mined by O and AB, draw 

 the perpendicular OC to 

 AB, then it is evident that 

 the required plane must 

 cut the line AB in the 

 point C. Since OC is the 

 only perpendicular which 

 can be drawn from O to 

 AB, then, by the previous 

 case, the problem is solved. 



Cor. Hence only one plane can be drawn 

 through a given point perpendicular to a given 

 line. 



PROP. IV. Through a given point to draw a 

 line perpendicular to a given plane. 



(i.) If the given point be on the plane. 



Let O be the given point, and P the 

 plane; it is required to 

 draw through O a line 

 perpendicular to the plane 

 P. Through O in the 

 given plane draw any line 

 AB, and through the 

 point O draw a plane Q 

 perpendicular to the line 

 AB by the last proposition, and in this plane 

 through the point O draw OD perpendicular to 

 OC. Then OD is the line required that is, it is 

 perpendicular to the given plane P. For 1 it is 

 perpendicular to OC, and also perpendicular to 

 OA ; therefore it is perpendicular to the plane of 

 these two lines that is, perpendicular to P. 



(2.) If the given point be without the plane. 



In the given plane P draw any straight line AB, 

 and from O draw a plane perpendicular to the 

 line intersecting the given 

 plane P in the line AC. 

 Then in the plane OAC, 

 from the point O draw 

 OD perpendicular to AC. 

 Then OD is perpendicular 

 to the given plane P, for 

 it is perpendicular to any 

 other line DB through its 

 foot in the plane. Let OD 

 be extended beyond the plane to O', so that O'D 

 is equal to OD. Join O and O' with A and B. 

 Comparing the two triangles OAB and O'AB, we 

 have the two sides OA, AB ( in the one equal to 

 the two O'A, AB in the other, and the contained 

 angles equal ; therefore the side OB is equal to 

 O'B. Then in the two triangles ODB, O'DB, 

 we have three sides in the one equal to three 

 sides in the other, and therefore the angle ODB 

 is equal to the angle O'DB ; each therefore is a 

 right angle ; therefore OD is perpendicular to 

 DB, and it is also perpendicular to DA 5 it is 

 therefore perpendicular to their plane P. 



PROP. V. Of the straight lines which can be 

 drawn from an external point to a plane, the per- 

 pendicular is the shortest; and of the others, those 

 whose extremities are equally distant from the 

 foot of the perpendicular are equal, and con- 

 versely. 



Let OA be drawn perpendicular to the plane P 



M 



