Let two planes AB, BC cut each other, and let B, D be two points in the line of their common section. Join BD. Because the points B, D are in the plane AB, the line BD is in the plane AB (Def. 8. 1); and because the points B, D are in the plane BC, the line BD is in the plane BC; therefore the line BD is common to the planes AB and BC, or it is the common section of these planes. Therefore, if two planes &c. Q. E. D.

PROPOSITION IV. THEOREM.

A

If a straight line be perpendicular to each of two straight lines in their point of intersection, it will also be perpendicular to the plane in which those lines are situate.

Let the line AB be perp. to each of the lines AC, AD in their point of intersection A; AB is also perp. to the plane passing through those lines.

Through A draw the line AE bisecting the angle CAD; and also draw any line AF. Take AC AD; draw CD meeting the lines AE, AF in the points E and F; join BC, BD, BE, BF.

Because the side AC is equal to AD, and AE bisects the angle CAD, the side CEDE, and AEC, AED are right angles (Cor. 10. 1). Consequently ADo = AE2 + DE' (B. 2). In the right angled triangles ABC, ABD the side AC = AD), and AB is common, therefore the side BC BD (2 Cor. B. 2).

B

C

F

E

A

D

Hence the triangles BCE, BDE are equal in all respects (8.1), therefore the angles BEC, BED are equal, and therefore are right angles. Consequently BD = BE+DE. From these equals take the former equals ADo = AÈo + DE3, then the AE AD9 BE remainders are equal, that is, BD2 AE, or AB BE AE; wherefore BAE is a right angle (1 Cor. B. 2), and AB is perp. to AE. Now AEF, BEF are right an

-

=

AE+

gled triangles, and therefore AF AE + EF2, and BF2 = BE2 + Eř2. Hence, by taking the former equals from the latter, BF2 AF2 BE2 - AE3, or BF2 -AF2 AB2; therefore BAF is a right angle, and AB is perp. to AF. But AF is any line drawn through the point A; therefore AB is perp. to the plane which passes through the lines AC, AD (Def. 1). Therefore, if a straight &c. Q. E. D. NULTY.

PROPOSITION V. THEOREM.

If three straight lines meet in one point, and a straight line be perpendicular to each of them in that point, those three lines are in one and the same plane.

Let the line AB be perp. to each of the three lines BC, BD, BE, at the point B where they meet; BC, BD, BE are in one and the same plane.

If not; let, if it be possible, BD and BE be in one plane, and BC above it: and let a plane pass through AB, BC; then the common section of this plane with the plane in which BD and BE are situate will be a straight line (3. 11). Let this line be BF; therefore AB, BC, BF are in one plane, which passes through AB, BC.

Because AB is perp. to each of the lines BD, BE, it is also perp. to the plane passing through them (4. 11), and therefore is perp. to every line meeting it in that plane. But BF meets AB in that plane; therefore ABF is a right angle. But, by the hypothesis, ABC is a right angle; therefore the angle ABF is equal to ABC; that is, because these angles are both in the same plane, a part is equal to the whole, which is impossible. Therefore BC is not above the plane in which BD and BE are situ

A

B

F

D

E

ate; wherefore BC, BD, BE are in one and the same plane. Therefore, if three &c. Q. E. D.

PROPOSITION VI. THEOREM.

If two straight lines be perpendicular to the same plane they will be parallel to each other.

A

C

Let the lines AB, CD be perp. to the same plane BDE; AB is parallel to CD.

Let the lines AB, CD meet the plane in the points B, D; and let B, D be joined. Because AB, CD are perp. to the same plane, they are also perp. to the line BD in that plane (Def. 1. 11); therefore they are parallel to each other (Cor. 28. 1). Q. E. D.

B

COR. If two straight lines be parallel, and if one of them be perp. to any plane, the other also will be perp. to the same plane.

PROPOSITION VII. THEOREM.

If two straight lines be parallel, and if one of them be perpendicular to a plane, the other also is perpendicular to the same plane.

Let AB, CD be two parallel lines, and let one of them AB be perp. to a plane EF; the other CD is perp. to the same plane.

For, if CD be not perp. to the plane EF to which AB is perp. let DG be perp. to it; then DG is parallel to AB (6. 11). Therefore DG and DC are both parallel to AB, and are drawn through the same point D, which is impossible (Ax. 10, p. 40). Therefore, if two &c. Q. E. D.

E

A

B

C G

PROPOSITION VIII. THEOREM.

F

Two straight lines which are parallel to the same straight line, and are not both in the same plane with it, are parallel to each other.

Let AB, CD be each of them parallel to EF, and not in the same plane with it; AB is parallel to CD.

In EF take any point G, from which draw, in the plane passing through EF, AB, the line GH perp. to EF; and in the

plane passing through EF, CD, draw GK perp. to EF. Because EF is perp. to GH and GK, it is perp. to the plane HGK passing through them (4. 2. Sup.). But EF is parallel to AB; therefore AB is perp. to the plane HGK (7. 2. Sup.). For

the same reason CD is at right angles to the plane HGK. Therefore AB, CD are perp. to the plane HGK; therefore AB is parallel to CD (6. 2. Sup.). Wherefore, two straight lines &c. Q. E. D.

PROPOSITION IX. THEOREM.

If two straight lines which meet each other be parallel to two other straight lines which meet each other toward the same parts, though not in the same plane with the first two, the first two lines and the other two will contain equal angles.

Let the two straight lines AB, BC, which meet each other, be parallel to the two straight lines DE, EF, which meet each other toward the same parts as AB, BC, and are not in the same plane with AB, BC; the angle ABC is equal to the angle DEF.

Take BA ED, and BC= EF; and draw AD, CF, BE, AC, DF. Because BA is equal and parallel to ED, AD is equal and parallel to BE (33. 1); and because BC is equal and parallel to EF, CF is equal and parallel to BE. Therefore AD and CF are equal and parallel to BE; therefore AD is equal and parallel to CF (8. 2 Sup.); therefore AC is equal and parallel to DF. Now because AB, BC are equal to DE, EF, and the base AC to DF, the angle ABC is equal to DEF (8. 1). Therefore, if two straight &c. Q. E. D.

D

B

PROPOSITION X, PROBLEM.

F

To draw a straight line perpendicular to a plane, from a given point above it,

Let A be the given point above the plane BH: it is required to draw from the point A a straight le perp to the paze bH. In the plane draw any straight line BC, and from the point A draw AD perp to BC: the AD be also perp to the plane BH the thing required is done. But if it be not. from the point D draw, in the pane BH, the straight line DE perp. to BC; and from the point A draw AF perp, to DE; and through F draw GH parallel to BC.

Because BC is perp. to EL and DA, it is perp. to the plane passing through ED, DA (4.11). But Gh is parallel to BC; wherefore G CH is perp. to the plane passing through ED, DA (7. 11), and is therefore perp. to every straight line meeting it in that plane (1 Def. 11). But AF, which

B

E

F

D

A

H

C

is in the plane passing through ED, DA, meets GH; there fore GH is perp. to AF. But AF is perp. to DE. Hence AF is perp. to each of the lines GH, DE; therefore AF is perp. to the plane BH, which passes through ED, GH. Therefore from the given point A above the plane BH the straight line AF is drawn perp. to that plane. Which was to be done.

COR. If it be required to erect a perpendicular to a plane from a point C in the plane, take a point A above the plane, and draw AF perp. to the plane; then from C draw a line parallel to AF, and it will be the perpendicular required (7. 11).

PROPOSITION XI. THEOREM.

If two parallel planes be cut by a third plane their sections with it are parallel lines.

For if the lines be not parallel, they must meet if produced; and if they meet, the planes in which they are situate must meet, which is impossible (14 Def. 1). Therefore, if two parallel &c. Q. E. D.

PROPOSITION XII. THEOREM.

If two straight lines be cut by parallel planes they will be cut in the same ratio.