Book XI By the fame reafon, FC will alfo be equal to FD. Again, because A D is equal to C B, and FA to FB; the two fides FA, AD will be equal to the two fides F B, BC, each to each; and the base FD has been proved to be equal to the bafe FC: Therefore the angle F A D [by 8. 1.] is equal to the angle FB C. Again, AG has been proved to be equal to BH: But FA alfo is equal to FB: Therefore the two fides FA, A & are equal to the two fides FB BH. And the angle FAG has been proved to be equal to the angle FBH: Therefore the base FG [by 4. 1.] is equal to the bafe F H. And again, because GE has been proved to be equal to E H, and E F is common; the two fides G E, EF, will be equal to the two fides HE, EF, and the bafe FH is equal to the base FG: Therefore the angle GEF is equal to the angle HEF [by 8. 1.], and fo the angles GEF, HEF are each a right angle: Therefore FE is at right angles to GH any how drawn thro' E. After the fame manner we demonftrate that F E is at right angles to all right lines that touch it, and are in the fame plane. But [by 3. def. 11.] one plane is perpendicular to another, when it is at right angles to all the right lines touching it, and being in the fame plane. Wherefore E F will be at right angles to the plane drawn through the right lines A B, C D.

If therefore a right line be at right angles to two right lines cutting one another in the common fection [of two planes;] it will be at right angles to the plane drawn thro thofe lines. Which was to be demonftrated.

If a right line stands at right angles to three right lines meeting one another in their common section; thofe three right lines will be all in the fame plane.

For let any right line AB ftand at right angles to the three right lines BC, BD, BE, in their point of contact B: I fay the right lines BC, BD, BE are all three in the fame plane.

For if not, let, if poffible, BD, BE be in one plane, and BC above it; and let the plane paffing through AB, BC be produced: Then the common section will make

[by

[by 3. 11.] a right line in the plane beneath, which let be B F. Therefore the three right lines A B, BC, BF are in one plane drawn thro' A B, BC. And because A B ftands at right angles to BD, BE, it will be at right angles to the plane drawn thro' BD, BE [by 4. 11.]. But the plane drawn thro' DB, BE is the plane beneath: Therefore a B is perpendicular to the plane beneath: Wherefore [by 3. def. 11.] it will be perpendicular to all right lines touching it, which are in the fame plane: But BF being in the plane beneath, touches it: Therefore ABF is a right angle. But the angle ABC is

A

-D

alfo fuppofed to be a right angle: Therefore the angle ABF is equal to the angle ABC; and they are both in the fame plane; which [by 9. ax.] is impoffible. Therefore the right line BC is not above the plane beneath: Therefore the three right lines BC, BD, BE are all in one plane.

Wherefore if a right line ftands at right angles to three right lines meeting one another in their common fection; those three right lines will be all in the fame plane. Which was to be demonftrated.

PROP. VI. THEOR.

If two right lines be at right angles to the fame plane, thofe right lines will be parallel to one another.

For let two right lines A B, C D be A at right angles to a piane: I fay A Bis parallel to CD.

For let them meet the plane in the points B, D; and join BD by a right line, and draw DE in the pine at right angles to BD, make DE equal to ab, and join BE, AE, AD.

Then because A D is perpendicular to the plane; it will be [by 3. det. II.] perpendicular to all rig plane which touch it, being both in the plane,

lies in that

But B D, BE,
touch AB:

C

B

D

Therefore

Book XII, Therefore the angles A B D, A BE are each of them right angles. Alfo by the fame reafon the angles CDB, CDE are each right angles: And because A B is equal to DE, and BD is common, the two fides A B, BD are equal to the two fides ED, DB, and they contain a right angle: Therefore [by 4. 1.] the base A D is equal to the base B E. And because A B is equal to DE, and AD to B E, the two fides A B, BE are equal to the two fides E D, D B, and their base AE is common: Therefore [by 8. 1.] the angle ABE is equal to ED A. But ABE is a right angle. Therefore alfo EDA is a right angle: And fo ED is perpendicular to DA. But it is also perpendicular to BD, DC: Wherefore ED ftands in the point of contact at right angles to the three right lines BD, DA, DC: Wherefore the three right lines B D, DA, DC [by 5. 11.] are in one plane. But A is in the fame plane wherein are BD, DA, for [by 2. 11.} every triangle is in one plane: Therefore AB, BD, DC are in one plane; and ABD, BDC are each right angles : Wherefore [by 28. 1.] AB is parallel to C D.

Therefore if two right lines be perpendicular to the fame plane; thofe right lines will be parallel. Which was to be demonftrated.

PROP. VII. THEOR.

If two right lines be parallel, and any points be taken in each of them; the right line which joins thofe points will be in the fame plane wherein are the parallels.

Let there be two parallel right lines A B, C D, and let any points E, F be taken in each of them: I fay the right line joining the points E, F is in the fame plane wherein are the parallels.

For if it be not, let it, A

C

E

G

if poffible, be elevated above

fpace, which [by 12. ax.] is impoffible: Therefore the right line drawn from the point E to F cannot be elevated above the plane paffing thro' the parallels AB, CD, consequently

it is in the fame.

Wherefore if two right lines be parallel, and any points be taken in them, the right line joining the faid points will be in the same plane wherein are the parallels. Which was to be demonftrated:

If there be two right lines parallel, and one of them be at right angles to fome plane, the other will also be at right angles to the fame plane.

Let there be two parallel right lines, AB, CD, and let one of them, as AB, be at right angles to fome given plane; the other CD will also be at right angles to that plane.

C

For let AB, CD meet the plane in the points B, D; and join BD: Then [by 7. 11.] AB, DC, DB are all in one plane. Draw DE in the given plane at right angles to BD, make D E equal to A B, and join BE, AE, AD: Then because AB is perpendicular to the given plane, it will be [by 3. def. 11.] perpendicular to all right lines which touch it, and are in the given plane: Therefore ABD, A BI are each of them right angles. But because the right line BD falls upon the two parallel right lines AB, CD; the angles ABD, CDB [by 29. 1.] will be equal to two right angles: But A B D is a right angle: Therefore CDB is alfo a right angles and so CD is perpendicular to BD. And because A B is equal to DE, and B D is common, the two fides AB, BD are equal to the two fides ED, DB and the angle ABD is equal to the angle EDB, for each of them is a right angle; therefore the base AD [by 4. 1.] is equal to the bafe BE: And because A B is equal to DE, and BE to AD, the two fides, A B, BE, will be equal to the two fides E D, DA, each to each, and their base AE is common: Wherefore

B

A

E

[by 8. 1.] the angle ABE is equal to the angle EDA: but ABE is a right angle; therefore EDA alfo is a right angle,

Y

and

and fo ED is perpendicular to D A. But it is alfo perpendicular to DB: Therefore [by 4. 11.] ED will be perpendicular to the plane drawn thro' BD, DA; and will be at right angles to all right lines which touch it, and are in that fame plane. But DC is in the plane drawn thro' в A, AD, because A B, BD [by 2. 11.] are in the plane drawn thro' BD, DA: But [by 7. 11.] A B, BD are in the fame plane as D C is: Therefore ED is at right angles to DC: and fo CD is at right angles to DE. But fo alfo is CD to BD. Therefore CD ftands at D, at right angles to two right lines DE, DB mutually cutting one another in the common fection of the two planes; and accordingly [by 4. II.] CD is at right angles to the plane drawn thro' DE, DB. But the plane drawn thro' DE, DB is the given plane: Therefore CD will be at right angles to the given plane. Which was to be demonftrated.

PROP. IX. THEOR.

Thofe right lines which are parallel to the fame right line, but not in the fame plane with it, are alfo parallel to one another.

For let A B, CD be each of them parallel to the right line EF, but not lying in the fame plane with it: I fay A B is parallel to CD.

For take any point G in EF, and from the fame draw GH in the plane paffing thro' E F, A B, at right angles to

A

E

H

K

G

B

D

F

EF; and again draw G K at right angles to EF in the plane paffing thro' FE, CD. Then because EF is perpendicular to GH, and to GK; EF [by 4. 11.] will also be at right angles to the plane paffing thro' GH, GK, and E F is parallel to A B: Therefore [by 8. 11.] A B is alfo at right angles to the plane paffing thro' H, G, K. By the fame reason CD is alfo at right angles to the plane paffing thro' H, G, K: Therefore A B, C D, each of them will be at right angles to the plane paffing thro' H, G, K. But if two right lines be at right angles to the fame plane; [by 6. 11.] they will be parallel to one another: Therefore AB is parallel to CD. Which was to be demonftrated.

PROP.