which is in the Plane paffing thro' ED and DA, doth touch it. Therefore GH is perpendicular to AF, and fo AF is perpendicular to GH; but AF likewife is perpendicular to DE; therefore AF is perpendicular to both HG, DE. But if a Right Line ftands at Right Angles to two Right Lines, in their common Section, that Line will be + at Right An-† 4 of this. gles to the Plane paffing thro' thefe Lines. Therefore AF is perpendicular to the Plane drawn thro' ED, GH; that is, to the given Plane BH. Therefore AF is drawn from the given Point A, above the given Plane BH, perpendicular to the faid Plane; which was to be done.

PROPOSITION XII.

PROBLEM.

To erect a Right Line perpendicular to a given ·Plane, from a Point given therein.

ET A be a given Point in a given Plane MN.

LE

It is required to draw a Right Line from the Point A, at Right Angles, to the Plane MN.

Let fome Point B be fuppofed above the given

Plane, from which let BC be drawn * perpendicular * 11 of this. to the faid Plane; and let AD be drawn + from A† 31. 1. parallel to BC.

Then because AD, CB, are two parallel Right Lines, one of which, viz. BC, is perpendicular to the Plane MN; the other AD fhall be ‡ also perpendi- † 8 of this. cular to the fame Plane. Therefore, a Right Line is erected perpendicular to a given Plane, from a Point given therein; which was to be done.

PROPOSITION XIII.

THEORE M.

Two Right Lines cannot be erected at Right Angles, to a given Plane from a Point therein given.

FOR

OR, if it is poffible, let two Right Lines AB, AC, be erected perpendicular to a given Plane on the fame Side, at a given Point A, in a given Plane.

Then let a Plane be drawn thro' BA, AC, cutting 3 of this. the given Plane thro' A in the Right Line DAE; therefore the Right Lines AB, AC, DAE, are in one Plane. And because CA is perpendicular to the + Def. 3. given Plane, it fhall also be † perpendicular to all Right Lines drawn in that Plane, and touching it; but DAE being in the given Plane, touches it. Therefore the Angle CAE is a Right Angle. For the fame Reafon, BAE is also a Right Angle; wherefore the Angle CAE is equal to BAE, and they are both in one Plane, which is abfurd. Therefore, two Right Lines cannot be erected at Right Angles, to a given Plane, from a Point therein given ; which was to be demonftrated.

PROPOSITION XIV. .

THEOREM.

Thofe Planes, to which the fame Right Line is per-
Fendicular, are parallel to each other.

ET the Right Line AB be perpendicular to each

For if they be not, let them be produced till they meet each other, and let the Right Line GH be the common Section, in which take any Point K, and join AK, BK. Then because A B is perpendicular to the Plane EF, it fhall also be perpendicular to the Right Line BK, being in the Plane EF produced. Wherefore the Angle ABK is a Right Angle. And for the fame Reafon, BAK is also a Right Angle.

And fo the two Angles ABK, BAK, of the Trian gle ABK, are equal to two Right Angles, which is impoffible. Therefore the Planes CD, EF, being* 17. Ip produced, will not meet each other, and fo are neceffarily parallel. Therefore, thofe Planes, to which the fame Right Line is perpendicular, are parallel to each other; which was to be demonstrated.

PROPOSITION XV.

THEOREM,

pa

If two Right Lines, touching one another, be rallel to two Right Lines, touching one another, and not being in the fame Plane with them, the Planes drawn thro' thofe Right Lines are parallel to each other.

ET two Right Lines AB, BC, touching one

Right EF,

touching one another, but not in the fame Plane with them. I fay, the Planes paffing thro' AB, BC, and DE, EF, being produced, will not meet each other.

For let BG be drawn from the Point B, perpendicular to the Planè paffing thro' DE, EF, meeting the fame in the Point G; and thro' G let GH be drawn parallel to ED, and GK parallel to EF; then because BG is perpendicular to the Plane paffing thro'. DE, EF, it fhall alfo make * Right Angles* Def. 3. with all Right Lines that touch it, and are in the fame Plane; but GH and GK, which are both in the fame Plane, touch it. Therefore each of the Angles BGH, BGK, is a Right Angle. And fince BA is parallel to GH, the Angles GBA, GBH, are †† 29. x. equal to the Right Angles: But BGH is a Right Angle; wherefore GBA fhall also be a Right Angle, and fo BG is perpendicular to BA. For the fame Reafon, GB is alfo perpendicular to BC. Therefore fince a Right Line BG, ftands at Right Angles to two Right Lines BA, BC, mutually cutting each other; BG fhall also be at Right Angles to the Plane drawn‡ 4 of this. thro' BA, BC. But it is perpendicular to the Plane drawn thro' DE, EF; therefore BG is perpendicujar to both the Planes drawn thro' AB, BC, and

DE,

DE, EF. But thofe Planes to which the fame Right 14 of this. Line is perpendicular, are parallel. Therefore the Plane drawn thro' AB, BC, is parallel to the Plane drawn thro DE, EF. Wherefore, if two Right Lines, touching one another, be parallel to two Right Lines, touching one another, and not being in the fame Plane with them, the Planes draum thro' these Right Lines are parallel to each other.

PROPOSITION XVI.

THEOREM.

If two parallel Planes are cut by any other Plane, their common Sections will be parallel.:

ET two parallel Planes, AB, CD, be cut by any Plane EFHG, and let their common Sections be EF, GH. I fay EF is parallel to GH.

For if it is not parallel, EF, GH, being produced, will meet each other either on the Side FH, or EG. First let them be produced on the Side FH, and meet in K; then becaufe EFK is in the Plane AB, all Points taken in EFK will be in the fame Plane. But K is one of the Points that is in EFK. Therefore K is in the Plate A B. For the fame Reafon K is alfo in the Plane CD. Wherefore the Planes AB, CD, will meet each other. But they do not meet, fince they are fuppofed parallel. Therefore the Right Lines EF, GH, will not meet on the Side FH. After the fame manner it is proved, that they will not meet, if produced, on the Side EG. But Right Lines, that will neither way meet each other, are pa tallel; therefore EF is parallel to GH. If, therefore, tava parallel Planes are cut by any other Plane, their common Sections will be parallel; which was to be demonítrated.

If two Right Lines are cut by parallel Planes, they hall be cut in the fame Proportion.

ET two Right Lines AB, CD, be cut by Parallel Planes GH, KL, MN, in the Points A, E, B, C, F, D. I fay, as the Right Line AE is to the Right Line E B, fo is CF to FD.

*

For let A C, BD, AD be joined: Let A D meet the Plane KL in the Point X, and join EX, XF. Then becaufe two parallel Planes KL, MN, are cut by the Plane EBDX, their common Sections EX, BD, are parallel. For the fame Reafon, becaufe 16 of thi two parallel Planes GH, KL, are cut by the Plane AXFC, their common Sections AC, FX, are parallel; and because EX is drawn parallel to the Side BD of the Triangle ABD, it fhall be as AE is to EB, fo is AX to XD. Again, because XF ist 2. 6. drawn parallel to the Side AC of the Triangle ADC, it fhall be + as AX is to XD, fo is CF to FD. But it has been proved, as AX is to XD, fo is AE to EB. Therefore, as AE is to EB, fo is t CF to 1.5 FD. Wherefore, if two Right Lines are cut by parallel Planes, they shall be cut in the fame Proportion; which was to be demonftrated.

PROPOSITION XVIII.

THE ORE M.

If a Right Line be perpendicular to fome Plane, then all Planes paffing thro' that Line will be perpendicular to the fame Plane.

ET the Right Line A B be perpendicular to the

are likewife perpendicular to the Plane CL.

For let a Plane DE país thro' the Right Line AB, whofe common Section, with the Plane CL, is the Right Line CE; and take fome Point F in CE; from