PROP. V. THEOR.

If one (AB, see fig. to prec. prop.), of two parallel right lines (AB and CD), be perpendicular to a plain (EF), the other (CD) is perpendicular to the same plain.

Draw in the plain EF, the right line BD, and in the same plain, DG perpendicular to BD, and equal to AB, and let BG, AG and AD be joined.

Because in the triangles BAD and BGD, the sides AB and DG are equal (Constr.), BD common, and the angles ABD and BDG equal, being each of them right, AD and BG are equal (4. 1 Eu.), whence the triangles AGB and AGD having also AB and DG equal, and AG common, the angle ADG is equal to ABG (8.1 Eu.), and therefore a right one; therefore DG being perpendicular to BD and AD, is perpendicular to the plain passing by them (2. 2. Sup.), in which plain is DC, since AB and DC are in the same plain (Hyp. and Def. 34. 1 Eu.), therefore DG is perpendicular to DC (Def. 3..2 Sup.); but AB and CD being parallels (Hyp.), and the angle ABD right (Hyp. and Def. 3. 2 Sup.), the angle CDB is also right (29. 1 Eu.); therefore CD being perpendicular to both BD and DG, is perpendi cular to the plain EF in which they are (2. 2 Sup).

PROP. VI. PROB,

On a given plain (AB), from a given point (C) not therein, to let fall a perpendicular.

Having drawn DE at pleasure in the plain AB, let fall thereon from the point C, the perpendicular CF A (12. 1 Eu.), in the plain AB by F draw GH perpendicular to DE (11. 1 Eu.), and let fall thereon from the point C the perpendicular CK, (12. i Eu.), which is the perpendicular required.

D

K

M

For through K having drawn LM parallel to DE (31. 1 Eu.), because DF is perpendicular to FC and FK (Constr.), it is perpendicular to the plain passing by them (2.2 Sup.), in which

plain is CK (Cor. Post. B. 2. 2. Sup.); whence LK, being parallel to DF (Constr.), is perpendicular to the same plain (5. 2 Sup.), and therefore the angle CKL is a right angle (Def. 3. 2 Sup.); whence, the angle CKH being also a right angle Constr.), CK is perpendicular to the plain passing by KL and KH (2. 2 Sup.), or to the plain AB.

PROP. VII. PROB.

To a given plain (AB), at a given point therein (C), to erect a perpendicular.

From any point D without the plain AB, let DE be drawn perpendicular to it (6. 2 Sup.), and through C draw CF parallel A to ED (31. 1 Eu.); and because ED is perpendicular to the plain AB (Constr.), CF, which is parallel to it, is perpendicular to the same plain (5. 2 Sup.), and therefore what was required is done.

PROP. VIII. THEOR.

F

CE

From the same paint in a plain, there cannot be two right lines at right angles to the plain, on the same side of it: and there can be but one perpendicular to a plain from a point above it.

A

D

B

For, if it be possible, let two right lines CD and CE be at right angles the same plain AB, from the same point C in the plain, and on the same side of it; and let a plain pass by CD and CE (Cor. Post. B. 2 Sup.), the common section of which with F the plain AB is a right line passing by C (1. 2 Sup.); let FCG be their common section, therefore the right lines CD, CE and FCG are in the same plain; and because CD is perpendicular to the plain AB (Hyp.), it is perpendicular to every right line drawn through C in that plain (Def. 3. 2 Sup.), and therefore to the right line FCG, therefore the angle DCG is a right angle; for a like reason ECG is a right angle; therefore the angles ECG and DCG are equal, part and whole, which is absurd. And from a point

above a plain, there can be but one perpendicular to the plain, for if there could be two, they would be parallel to each other (5. 2. Sup.), which is absurd.

PROP. IX. THEOR.

If a right line (AB) be perpendicular to a plain (CD), all the plains drawn thereby are perpendicular to the same plain.

Let EF be a plain passing by AB, and EG the common section thereof with the plain CD, from any point wherein H in the plain EF, let HK be drawn perpendicular to EG (11. 1 Eu.), which, the angles ABH and KHB being each of them right E angles, is parallel to AB (28. 1 Eu.), and therefore perpendicular to the plain CP (Hyp. & 5. 2 Sup.); the like might be proved of any right line drawn

in the plain EF perpendicular to the common section EG, therefore the plain EF perpendicular to the plain CD (Def. 4. 2 Sup.); the like might be proved of any other plain passing by AB.

Cor. If from the point (B), wherein a perpendicular (AB) to a plain (CD) meets it, a perpendicular (BG) be drawn to any right line (DL) in the same plain; a right line (AG), joining any point (A) in the perpendicular to the plain, to the point (G) where the perpendicular to the right line meets it, is perpendi cular to the same right line (DL).

In the right line DL take equals GD and GL, and join BD, BL, AD and AL; the triangles BGD and BGL right angled at G, having BG and GD severally equal to BG and GL, the right lines BD and BL are equal (4. 1. Eu.); whence, the triangles ABD and ABL having AB common, and the angles at B right (Hyp. and Def. 3. 2 Sup.), AD and AL are equal (4. 1 Eu.); whence, the triangles AGD and AGL having AG common, and GD equal to GL, the angles AGD and AGL are equal (8. 1 Eu.), and therefore right (Def. 20. 1 Eu).

PROP. X. THEOR.

If two plains (AB and CB), intersecting each other, be perpendicular to a third (ADC); their common section (DB) is perpendicular to the same plain.

If DB be not perpendicular to the plain ADC, draw from the point D in the plain AB, the right line DE perpendicular to AD (11.1 Eu.); and in the plain CB, from the point D, draw DF perpendicular to CD (by the same); and because the plain AB is perpendicular to the plain ADC, and DE is drawn in the plain AB, perpendicular to AD their common section, DE is perpendicular

H

to the plain ADC (Def. 4. 2 Sup). In like manner DF may be proved to be perpendicular to the plain ADC. Therefore two right lines DE and DF are at right angles to the same plain ADC, on the same side of it, at the same point D, which is absurd (8. 2. Sup.). Therefore there cannot be any right line perpendicular to the plain ADC at the point D, except DB; therefore DB is perpendicular to the plain ADC.

PROP XI. THEOR.

Two right lines (AB and CD), parallel to the same right line (EF), which is not in the same plain with them, are parallel to each other.

A

E

B

G

D

From any point G in EF, let two perpendiculars GH and GK to EF be drawn (11. 1 Eu.), one in the plain passing by EF and AB, and the other in that passing by EF and CD, meeting AB and CD in H and K; and EF being perpendicular to GH and GK (Constr.), is perpendicular to the plain passing by them (2. 2 Sup.); whence AB and CD being parallel to EF, are perpendicular to the same plain (5. 2 Sup.), and therefore parallel to each other (4. 2. Sup).

C

PROP. XII. THEOR.

If two right lines (AB and AC), meeting each other (as in A), be parallel to two others (DE and DF), likewise meeting each other (as in D), though not in the same plain with them; the first two and the last two contain equal angles.

Let AB and DE be taken equal to each other, and also AC and DF, and let BC, EF, BE, AD and CF be drawn. And since AB and DE are equal and parallel, BE and AD are equal and parallel (33. 1 Eu.); in like manner CF may be shewn to be equal and parallel to AD; therefore BE and CF are equal and parallel to each other (Ax. 1. 1 and 30. 1 Eu.), and therefore BC and EF are equal (33. 1 Eu.); whence the triangles ABC and DFF being mutuan, equilateral, the angles BAC and ED are equal (8. 1 Eu).

PROP. XIII. THEOR.

Plains (AB and CD), to which the same right line (EF) is perpendicular, are parallel.

For if not, let them, being produced, meet, and let their common section be the right line GH, to any point wherein as K, having drawn EK and FK, the angles KEF and KFE are each of them right angles Hyp. and Def. 3. 2 Sup.), and so two angles of the triangle KEF are equal to two right angles, which is absurd (17. 1 Eu.); therefore the plains AB and CD being produced, do not meet each other, and are therefore parallel (Def. 7. 2 Sup).