Then because two parallel planes are cut by the plane ACF in the lines BG, CF,

BG is parallel to CF,

and therefore AB: BC :: AG: GF.

(Th. IV.)

(Pl. G. IV. 15.)

In like manner because two parallel planes are cut by the

plane AFD in the lines GE, AD,

GE is parallel to AD,

and therefore AG: GF:: DE : EF.

Hence, the ratios AB: BC and DE:EF being each equal to the ratio AG: GF,

AB BC: DE : EF.

Q. E. D.

EXERCISES.

I. All lines joining two points in two parallel straight lines lie in the plane of the parallels.

2. If three parallel or three concurrent lines meet the same straight line, they lie all in the same plane.

3. If two planes are parallel to a given line, their intersection is parallel to the same line.

4. If three lines lie two and two in three planes, they are either concurrent or parallel.

5. If any number of lines diverge from the same point, the planes through any two pairs of these lines intersect in lines through the same point.

6. If any number of lines are parallel, the intersections of the planes through any two pairs of these lines are parallel

to them.

7. If a plane intersect a series of planes all passing through a given line, the lines of intersection all pass through the point where the given line meets the plane, or are parallel to the given line, if it does not meet the plane.

8. Planes through a given point and each of a series of parallel lines intersect another plane in lines passing through the point, where that line of the series which passes through the given point meets the plane.

(If the given point is the eye and the plane the picture plane, the point in the plane is the vanishing point of the perspective representation of the parallel lines.)

If from the extremities of two parallel lines AB, CD, parallel lines Aa, Bb, Cc, Dd are drawn to meet a plane in a, b, c, d respectively,

ab cd: AB : CD.

IO. If in the figure of Th. VII. CD be drawn meeting the plane BGE in K, the figure BKEG is a parallelogram : and the intersection of the diagonals of the parallelogram lies on the line joining the middle points of AD and CF.

=

II. P, p are two points in the line AB such that AP Bp through P, p two parallel planes pass meeting a line AC in Q, q, and a line BD in R, r: prove that the triangles PQR, pqr are equal.

I2. If parallel lines drawn through the angular points A, B, C, ... of a plane polygon meet a plane parallel to its plane in a, b, c, ... respectively the polygon abc... is identically equal to ABC.

13. If lines drawn from a point O through the angular points A, B, C... of a plane polygon meet a plane parallel to

its plane in a, b, c... respectively; the polygon abc... is similar to ABC..., and the ratio of corresponding sides is that of Oa to OA.

14. If in the preceding question the plane abc... is not parallel to ABC..., the corresponding sides ab and AB, be and BC, ... meet in points lying on the intersection of the two planes.

15. If the triangles ABC, abc in different planes are such that AB and ab meet (when produced), as also BC and bc, CA and ca; the lines Aa, Bb, Cc meet in one point, or are parallel.

2

16. АВ11, ABC, are shadows on the same plane of a triangle ABC cast by two bright points at 01, 0, the intersections of BC and B2C2, of C1A, and CA, and of A1B1, and A,B2 lie in one straight line, and AA, B1В С1С2 meet in one point.

2,

SECTION II.

NORMALS AND OBLIQUES.

THEOREM VIII.

There is one point in a plane, and only one, whose distance from a given point without the plane is less than that of any other point, and the straight line drawn from the given point to this point is perpendicular to every straight line through it in the plane.

Let O be a point without a given plane, and P any point in the plane.

Then, since O is not in the plane,

OP cannot be less than a certain finite length.

If A be a point in the plane such that OA is equal to this length, there is no position of P, for which OP is equal to OA:

for if there were, joining AP in the plane, OAP would be an isosceles triangle,

and therefore the angle OAP an acute angle:

therefore the perpendicular from O on AP would be less than OA. (Pl. G. I. 15.)

of

Therefore there is one point A in the plane, and only one, whose distance from O is less than the distances from all other points in the plane.

Let PAQ be any line, passing through A, in the plane,

then OA, being the shortest distance from A to the plane, is the least distance from O to the line PAQ;

therefore OA is perpendicular to the line PAQ,

(Pl. G. I. 15.) so that OA is perpendicular to every line in the plane passing through A. Q. E. D.

Def. 5. A straight line perpendicular to every straight line that meets it in a given plane is said to be normal to the plane, and the plane normal to the straight line.

COR. 1. (Euc. XI. 13). From a point outside a plane one, and only one, normal can be drawn to the plane.