But we may demonstrate it thus. If EF the common Section be not a right Line, let there be drawn in the Plane CD the right Line EOF, and in the Plane A B the right Line EQ F. The two right Lines therefore EOF, EQF, will include a Space. Which is abfurd.

PROP. IV. Theorem.

143

IF a right Line (BA) be perpendicular to two right Fig. 12. Lines (CAX, FA S) which cut each other, it will also be perpendicular to the Plane which is drawn thro them.

If you deny it, let another right Line BQ be perpendicular to the Plane of the right Lines AC, AF. Join AQ, and to this in the Plane FAC draw the Perpendicular QO. This being produced, will neceffarily cut (as is gather'd from Schol. Prop. 31. 1. 1.) one of the right Lines CAX, FAS, or both, wheresoever the Point Q shall be. Therefore let it cut CAX in O, and let BO be join'd. Because therefore the Angle BAO is by the Hypothefis a right one;

The Square of BO shall be equal to

BA Squ.
+
A O Squ.S
perpendicular to the

But because BQ is suppos'd Plane F A C, and consequently (by Defin. 3. l. 11.) makes a right Angle with AQ;

BA Squ. is equal to

BQ Squ.
+
AQ

(b)

And because the Angle A QO is by the Construction a right one;

AO Squ. is equal to

Squ.

(a) Per 47. 1.1.

(b) Per 47 1. 1.

OQ Squ.
+
AQ

2

Therefore BO Squ. is equal to BQ Squ.+.

OQ Squ.+ AQ Squ. twice taken.

There

144

Fig. 13.

Fig. 14.

Therefore BO Square is greater than the Squares of BQ and OQ; and (as is clear from Prop. 47. 1. 1.) consequently BQO is not a right Angle. Therefore BQ is not perpendicular to the Plane (by Defin. 3. 1. 11.) CAF. Therefore the Proposition is manifest.

Scholium.

FRom its being fuppos'd that B

Qis perpendicular to

the Plane FA C; it is directly demonftrated that it is not perpendicular to that Plane; and consequently from the denial of the Affertion of the Theorem, the fame Affertion is directly proved. This Demonstration, as to the Substance of it, is John Cierman's.

I

PROP. V. Theorem.

Fthree right Lines (BA, CA, FA) be perpendicular to the same right Line (AR) at the same Point (A); those three will be in one Plane.

For, if it may be, let one of them BA be in another Plane (RO) which may cut LQ the Plane of the other two CA, FA, in the right Line AO. Because by the Hypothesis R A stands perpendicularly upon the two CA, FA, it will be perpendicular to the Plane LQ (by the foregoing). Therefore RA makes a right Angle with AO (by Defin. 3. 1. 11.) But also by the Hypothefis RAB is a right Angle. Therefore the Angles RAB and RAO are equal. Which is absurd.

R

PROP. VI. Theorem.

are perpendicular to

Ight Lines (AB, CD) which
the Same Plane (CF) are parallel.

It might be taken for granted as a Thing of it felf

known; but we may demonstrate it thus.

BD being join'd, make in the Plane FE the Line

DG perpendicular to BD, and equal to BA; and let

DA,

•

Conftruction

DA, GA, GB, be be join'd. The right Lines BD, DG, are equal to BD (a) and BA; and the Angles BDG, (a) By the (b) DBA are right ones. Therefore (per. 4. 1. 1.) A D, (b) Per Def. BG, are equal. Therefore the Triangles ABG, GDA 3. 1. 11. are equilateral to each other, and consequently the Angles ABG, ADG are equal. But ABG (by Defin. 3.1. 11.) is a right Angle. Wherefore ADG is alfo a right one. But BDG also by the Construction, and CDG by Defin. 3. are right Angles. Therefore GD is perpendicular to three Lines CD, AD, BD. Therefore CD is (c) in one Plane with AD, and BD. But (c) By the A B also is in one Plane (per 2. 1. 11.) with AD and foregoing B D. Therefore A B, CD are in one Plane. Therefore feeing the Angles ABD, CDB (by Defin. 3. 1. 11.) are right ones, AB, CD will (per 29. 1. 1. and Defin. 36. 1. 1.) be parallel Lines. 2. E. D.

PROP. VII. Theorem.

A Line

cutting right Lines (AB, CD) Fig. 15.

placed in the Same Plane, is in one and the fame

Plane with them.

It might be taken for granted. But he that will may thus demonstrate it.

Let another Plane cut the Plane of the right Lines AB, CD, in the Points EF, If now EF is not in the Plane of A B, CD, it is not the common Section. Let EGF therefore be so. Therefore (per 3. 1. 11.) EGF is a right Line; the two right Lines therefore EF, EGF inclose a Space. Which is absurd.

Corollary.

For

HEnce it follows, that if EF cut the Parallels AB
CD, it is in the same Plane with them.
(by Defin. 36. 1. 1.) any two Parallels are in the same
Plane.

Fig. 14.

I

1

Fig. 16.

PROP. VIII. Theorem.

Fof two Parallels (AB, CD) one (AB) be perpendicular to a Plane (EF); the other alfo (CD) will be perpendicular to the fame Plane.

It might be taken for granted. If the Demonstrati on be requir'd, it is as follows.

[BD, AD being drawn; in the Plane EF make GD perpendicular to B D. It will also (fee the DemonStration of Prop. 6. 1. 11.) be perpendicular to AD Therefore (per 4. 1. 11.) GD will be perpendicular to the Plane ABD, that is (by the foregoing Coroll.) to the Plane CBDA. Wherefore (per Def. 3. 1. 11.) CDG is a right Angle. But the Angle CDB is also a right one; forasmuch as with ABD which (per Defin. 3. 1. 11.) is a right Angle, it maketh two right ones (per 27. 1. 1.) Therefore (per 4.1. 11.) CD is perpendicular to the Plane GDB or EF. Q. E. D.]

R

PROP. IX. Theorem.

i

IGHT Lines (AB, EF) which are parallel to the Same right Line (CD) altho they be not in the fame Plane with it, are also parallel betwixt themselves.

Altho it might be taken for granted, yet we will demonftrate it thus.

In the Plane of the Parallels A B, CD, draw GK perpendicular to CD. Likewise in the Plane of the Parallels EF, CD, draw HK perpendicular to CD. There(a) Per4.1.11, fore, (a) CK is perpendicular to the Plane GKH. Therefore, feeing AG, EH, are parallel to CK, the (b) Per8.1.11. fame AG, EH (b) will be perpendicular to the Plane (c) Per.6.1.11.G.KH. Therefore AG, EH (c) are parallel. 2. E. D.

PROP

4

I

PROP. X. Theorem.

F two right Lines (AC, BC) be parallel to two Fig. 17 right ones (DF, EF); albeit they be not in the Jame Plane, they comprehend equal Angles (Cand F).

Let CA, CB, be made equal to FD, FE, and let DE, AB, DA, FC, EB be drawn. Seeing AC, FD are parallel and equal, AD alfo and C F will (a) be pa- (a) Per 33 rallel and equal. In like manner I might shew BE, CF. to be parallel and equal. Therefore AD, BE, are alfo parallel (b) and equal (per Axiom. 1.) Therefore (per (b) By the 33. l. 1.) AB, DE, are equal. Seeing therefore the foregoing Triangles BA C, EDF are equilateral to each other,

the Angles C and F (c) are equal. 2. E. D.

O draw

To drau

(c) Per 8.1.1

a Perpendicular to a given Plane (AB) Fig. 18 Point given without it (C.)

The Construction. In the Plane AB draw any right Line as DF, unto which, from C erect the Perpendicular CE. Then in the Plane AB thro' E draw A EM perpendicular to the same DF. Then to AM from C draw the Perpendicular CG. I say that CG is perpendicular to the Plane, AB.

Thro' Glet HG be drawn parallel to DF. By the Construction D E is perpendicular to CE and EM. Therefore D E is perpendicular to the Plane EM (d), as alfo (d) Per 4.1. is HG (e). Therefore (by Defin. 3. 1. 11.) CG is perpen-(e) Per 8. dicular to HG. But CG by the Construction is also per-1.11. pendicular to E M. Therefore (f) CG is perpendicular (f) Per 4. to the Plane A B. Which was the Thing propos'd.

1. ii.

[Scholium. In Practice thus. Let there be a Cord Fig. 20. L.12

or Rule fastned to the given Point A: And from the same, let there be describ'd by the end of it B in the Plane given the Circle BCFL. The Line AK which connects the given Point and the Centre of the Circle, will be perpendicular to the given Plane: