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PROP. X: THEOR. If two right lines touching one another, be' parallel

to two right lines touching one another, but not being in the same plane; those right lines will con, tain 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 same plane: I say the angle A B C is equal to the angle DEF.

For take BA, BC, ED, EF equal to one another; and join AD, CF, BE, AC, DF. Then because B A is equal and parallel to ED; [by 35. 1.] AD will be equal and parallel to be. By the same reason CF will be equal and parallel to BE. Therefore AD, CF are each of them equal and parallel to BE.

B But right lines which are parallel to the same 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 A C, DF join them: Wherefore AC is equal and parallel to DF. And

E because the two fides A B, BC are equal to the two sides DE, EF,

and the D

F hale Ac is equal to the base D F; 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 same plane, those right lines will contain equal angles. Which was to be demonstrated.

PROP. XI. PROBL. From a given point above a given plane, to draw a

right line perpendicular to that plane '. 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 BC, and (by 12. 1.] from a draw AD perpendicular to B C. Y 2

If

If therefore ad be also perpendicu-
lar to the given plane; what was

required will then be done : But if
H

it be not, [by II. 1.) draw de in
the given plane from the point d'y
perpendicular to Bc; and (by 12.
1.) draw A F from the point A,

perpendicular to DE; and thro's
D

draw G H parallel to BC. Then because B C is at right angles to AD, DE; [by 4 11.) Be will be at right angles to the plane passing thro! LD, DA, and GH 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 also 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 same plane. But the right line A F touches G H being in the plane passing thro' ED, DA: Therefore g h is perpendicular to FA: Consequently FA is perpendicular to GĦ: But [by construction) a F also is perpendicular to De: Therefore AF is perpendicular to cH, and to de. But if a right line stands at right angles to two right lines touche ing one another in the common section; (by 4. 11.) it will be at right angles to the plane drawn thro' them: Wherefore F A is at right angles to the plane drawn thro' ED, GH. But the plane drawn thro' ED, GH is the given plane : Therefore ai is perpendicular to this plane. Wherefore a right line af is drawn from a given point А, A above a given plane, perpendicular

to that plane. Which was to be done.

1 The praćtice of this problem may be thus, From the given point A, draw three right lines & B; A C, A G, of the same length, meeting the plane in the points, B, C, G, which may be easily done, with a string or a pair of com. passes. Join the points B, C; C, 6; and Having bisected BC, ca in D, 3 draw the perpendiculars DË, E F to BC, ca,

meeting in the point y: Then will the F

right line ar be perpendicular to the B

plane B G. The demonftration being Geasy, I fall omit.

PROP

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PROP. XII. PROBL. A plane being given, and a point given in it; to

ere&t a right line from that point at right' angles to tbat planem.

Let A be the given point in a given plane: It is required to erect a right line from that point ac right anglés to that plane. Conceive some point B to be above

D the plane, and (by 11. 11.] draw from it the perpendicular B ç to the given plane: And [by 31. 1.) thro' A draw A D perpendicular to BC.

Then because the two right lines AD, CB are parallel, and Bc one of them is at right angles to the given

A Ć plane, the other AD [by 8. 11.) will be also 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 che plane. Which was to be done.

* This problem is eally resolved by help of a square, or a parallel ruler.

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PROP. XIII. THEOR.

At a point given in a given plane, two right lines,

cannot be erected perpendicular to the plane, on the fame siden.

For if it be possible, at the point a in the given plane, det two right lines A B, A C be drawn at right angles on the same fide; and thro' AB, AC draw a plane, which {by 3.11.] will make a right line in the given plane drawn

throA: which let be DAE; then IC B В

the right lines A B, AC, DAE, are in one plane. But because ca is at

right angles to the given plane, (by DA

3. def. 11.) it will be at right angles A

to all right lines in that plane which, E touch it. But the right line D A F, Y

3

being

being in the given plane, touches CA: Therefore CA E is a right angle : by the same reason B A E is also a right angle: Wherefore the angle CA E is equal to B A E, and they are both in one plane. Which [ by 9. ax.) is impossible.

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

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

PROP, XIV. THEOR. Tbose planes to which the same right line is perpendi

cular, are parallel to one another. For let the right line A B be perpendicular to each of the planes' C D, EF: I fay those planes are parallel.

For if they be not parallel, they will meet when produced. Let them be produced, and [by 3. 11.) their

common section is a right line, which let be GH. Take any point

K in GH, and join AK, B K. K

Then because AB is perpendicular to the plane, E F; [by 3. def. 11.]

it will also be perpendicular to the Fright line BK which is in the

plane E F produced: Wherefore А

B A B K is a right angle. By the

same reason B A K is also a right

angle; and so the two angles D E ABK, BAK of the triangle ABK,

are equal to the two right angles ABK, BAK, which [by 17. 1.) is impossible: Therefore the planes C D, E F produced will not meet. Wherefore the planes C D, E F are parallel.

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

H Η

1

PROP

PRO P. XV. THEOR. If two right lines touching one another, be parallel

to two right lines touching one another, but not in the same 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, E F touching one another, but not in the same plane: I say the planes which pass thro' A B, BC, DE, E F if produced will not meet.

For [by 11. 11.) draw BG from the point B perpendicular to the plane passing thro' D E, EF, meeting the plane in g; and thro' G [by 31. 1.] draw G H parallel to ED, and Gk to EF. Then because BG is perpendicular to the plane

E

I passing thro' DE, E F, and (by 3. def. 11.] it will be at right B

ra K к angles to all the right lines that touch it, and are in that plane; and GH, GK, which are both in the plane passing thro' DE, A E F, do touch it: Therefore the

DH angles BGH, BGK are each a right angle. And because B A is parallel to GH, the angles GBA, BGH [ by 29. L.) are equal to two right angles. But BGH is a right angle; therefore G B A will also be a right angle; and so G B is perpendicular to BA. By the same reason B G is also perpendicular to BC. Therefore because the right line BG stands at right angles to the two right lines B A, BC cutting one another: [by 4. 11.] BG also is perpendicular to the plane passing thro' A B, BC. [* and by the same reason 6G is perpendicular to the plane passing thro' GH, GK. But the plane paffing thro' GH, G K, is the same as that passing thro' DE, EF: Wherefore BG is perpendicular to the plane passing thro' DE, EF. But it has been proved that BG also is perpendicular to the plane passing thro' A B, B c.] and it is perpendicular to the plane passing thro' DE, EF: Therefore BG is perpendicular to both the planes passing thro' AB,

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

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