PROP. X: THEOR.

If two right lines touching one another, be parallel to two right lines touching one another, but not being in the fame plane; thofe right lines will contain equal angles.

For let two right lines A B, BC touching one another, be parallel to two right lines DE, EF touching one another but not in the fame plane: I fay the angle A B C is equal to the angle DEF.

B

For take BA, BC, E D, EF equal to one another; and join AD, CF, BE, AC, DF. Then because BA is equal and parallel to ED; [by 35. 1.] A D will be equal and parallel to BE. By the fame reafon CF will be equal and parallel to BE. Therefore AD, CF are each of them equal and parallel to BE. But right lines which are parallel to the fame right line, but not in the A fame plane [by 9. 11.] will be parallel to one another: Therefore A D is parallel and equal to CF; and the right lines AC, DF join them: Wherefore AC is equal and parallel to D F. And because the two fides AB, BC are equal to the two fides DE, E F, and the D

E

bafe AC is equal to the base DF; the angle ABC [by 8. 1.] will be equal to the angle DEF.

Therefore if two right lines touching one another be parallel to two right lines touching one another, but not being in the fame plane, thofe right lines will contain equal angles. Which was to be demonstrated.

PROP. XI. PROB L.

From a given point above a given plane, to draw a right line perpendicular to that plane 1.

Let A be a given point above a given plane: it is required to draw a right line from the point a, perpendicular to that given plane..

Draw any how in the given plane, the right line B C, and by 12. 1.] from A draw AD perpendicular to BC.

B

D

H

If therefore AD be alfo perpendicular to the given plane; what was required will then be done: But if it be not, [by 11. 1.] draw DE in the given plane from the point D, perpendicular to BC; and [by 12. 1.] draw A F from the point A, perpendicular to DE; and thro' F draw GH parallel to BC.

Then because B C is at right angles to AD, DE; [by 4: 11.] BC will be at right angles to the plane paffing thro ZD, DA, and G H is parallel to it. But if there be two parallel right lines, one of which is at right angles to fome plane, the other [by 8. 11.] will alfo be at right angles to that plane: Wherefore GH is at right angles to the plane paffing thro' E D, DA; and fo [by 3. def. 11.] it is perpendicular to all right lines touching it, which are in the fame plane. But the right line AF touches G H being in the plane paffing thro' ED, DA: Therefore G H is perpendicular to FA: Confequently FA is perpendicular to G H. But [by conftruction] AF alfo is perpendicular to DE: Therefore AF is perpendicular to H, and to DE. But if a right line ftands at right angles to two right lines touching one another in the common fection; [by 4. 11.] it will be at right angles to the plane drawn thro' them: Wherefore FA is at right angles to the plane drawn thro' ED, GH, But the plane drawn thro' ED, GH is the given plane: Therefore AF is perpendicular to this plane.

B

Wherefore a right line AF is drawn from a given point A above a given plane, perpendicular to that plane. Which was to be done.

A

The practice of this problem may be thus. From the given point a, draw three right lines A B, AC, A G, of the fame length, meeting the plane in the points, B, C, G, which may be eafily done, with a ftring or a pair of compaffes. Join the points B, C ; c, e; and having bifected в C, CG in D, E draw the perpendiculars DF, EF to BC, CA, meeting in the point F: Then will the right line a F be perpendicular to the plane B G. The demonftration being Geafy, I fhall omit.

PROP.

PROP, XII. PROBL.

A plane being given, and a point given in it; to erect a right line from that point at right anglès to that planem.

Let A be the given point in a given plane: It is réquired to erect a right line from that point at right anglés to that plane.

Conceive fome point B to be above the plane, and [by 11. 11.] draw from it the perpendicular B C to the given plane: And [by 31. 1.] thro' A draw A D perpendicular to BC.

Then because the two right lines A D, C B are parallel, and в Ç one of them is at right angles to the given

D B

АС

plane, the other AD [by 8. 11.] will be alfo at right angles to the given plane.

Therefore a plane being given, and a point given in it, a right line is erected from that point at right angles to the plane. Which was to be done.

This problem is easily refolved by help of a square, or parallel ruler.

PROP. XIII. THEOR.

At a point given in a given plane, two right lines, cannot be erected perpendicular to the plane, on the fame fide".

For if it be poffible, at the point A in the given plane, Jet two right lines A B, AC be drawn at right angles on the fame fide; and thro' A B, AC draw a plane, which [by 3.11.] will make a right line in the given plane drawn

thro' A: which let be DAE: then the right lines A B, AC, DA E, are in one plane. But becaufe CA is at right angles to the given plane, [by 3. def. 11.] it will be at right angles to all right lines in that plane which touch it. But the right line D A E, Y 3 being

Book XI. being in the given plane, touches CA: Therefore CA E is a right angle: by the fame reafon B A E is also a right angle: Wherefore the angle CAE is equal to B A E, and they are both in one plane. Which [by 9. ax.] is impoffible.

Therefore at a given point in a given plane, two right lines cannot be erected at right angles on the fame fide. Which was to be demonftrated.

"Tho' two right lines cannot be drawn from the fame point in a plane perpendicular to that plane on the fame fide of it, yet two fuch perpendiculars may be drawn, the one on one fide, and the other on the other fide of that plane.

PROP, XIV. THEOR.

Thofe planes to which the fame right line is perpendicular, are parallel to one another.

K

For let the right line AB be perpendicular to each of the planes CD, EF: I fay thofe planes are parallel. For if they be not parallel, they will meet when produced. Let them be produced, and [by 3. 11.] their common fection is a right line, which let be GH. Take any point K in GH, and join AK, BK. Then because AB is perpendicular to the plane. EF; [by 3. def. 11.] it will also be perpendicular to the F right line BK which is in the plane E F produced: Wherefore ABK is a right angle. By the fame reason B A K is also a right angle; and fo the two angles ABK, BAK of the triangle ABK, are equal to the two right angles

C

A

H

DE

B

Wherefore

ABK, BAK, which [by 17. 1.] is impoffible: Therefore the planes CD, EF produced will not meet. the planes CD, E F are parallel.

Therefore those planes to which the fame right line is perpendicular, are parallel. Which was to be demonftra

ted.

PROP.

PROP. XV. THEOR.

If two right lines touching one another, be parallel to two right lines touching one another, but not in the fame plane; the planes that pass thro' them will also be parallel.

For let two right lines A B, BC touching one another be parallel to two right lines DE, EF touching one another, but not in the fame plane: I fay the planes which pafs thro' A B, BC, DE, E F if produced will not meet.

For [by 11. 11.] draw BG from the point в perpendicular to the plane paffing thro' D E, EF, meeting the plane in G; and thro' G [by 31. 1.] draw GH parallel to ED, and GF to EF. Then because BG is perpendicular to the plane paffing thro' DE, E F, and [by 3. def. 11.] it will be at right B angles to all the right lines that touch it, and are in that plane;

E

F

G

K

and GH, GK, which are both in the plane paffing thro' D E, EF, do touch it: Therefore the angles BGH, BG K are each a

A

DH

right angle. And because в A is parallel to GH, the angles GBA, BGH [by 29. 1.] are equal to two right angles. But BGH is a right angle; therefore G B A will alfo be a right angle; and fo G B is perpendicular to BA. By the fame reason B G is alfo perpendicular to BC. Therefore because the right line B G ftands at right angles to the two right lines BA, BC cutting one another: [by 4. 11.] BG alfo is perpendicular to the plane paffing thro' A B, BC. [* and by the fame reafon BG is perpendicular to the plane paffing thro' GH, G K. But the plane paffing thro' GH, GK, is the fame as that paffing thro' DE, EF: Wherefore B G is perpendicular to the plane paffing thro' DE, EF. But it has been proved that BG alfo is perpendicular to the plane paffing thro' AB, BC.] and it is perpendicular to the plane paffing thro' DE, EF: Therefore BG is perpendicular to both the planes paffing thro' A B,

*The words within the braces, ought to be omitted.