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Again, :: ABCDE; BE = AD, :: AB, BE = AD, DE, each to each, and a E is common to the AS A BE, ADE. :. ABE = 2 ADE; but A B E is a right Z,.. ADE is a right Z, and Ed is DA: but Ed is 1 to BD and Dc: i. Ed is I to each of the three straight lines, DB, DA, DC, in the point in which they meet. :. they are in the same plane; but A B is in the plane of BD, DA, :: three straight lines which meet are in the same plane, :: AB, BD, DC, are in one plane; and each of the A B D, BDC, is a right 2 :: A B is parallel to cd.

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If two straight lines be parallel, the straight line drawn from

any point in the one to any point in the other is in the same plane with the parallels.

Let A B, CD, be parallel straight lines: take any point e in the one, and any point r in the other: the straight line joining

E and F shall be in the plane of the parallels.

For if not, let it, if possible, be above the plane, as EGF: and in the plane ABCD of the parallels :

draw the straight line EHF from E to F: and :: EGF is also a straight line, the two lines EGF, EHF, include a space, which is impossible. Therefore the straight line joining E and F is not above the plane ABCD; similarly it cannot be below it: it is ::. in the plane.





If two straight lines be parallel, and one of them be perpen

dicular to a plane: the other shall be also perpendicular to the same plane.

Let A B, CD, be the two parallels, and ABI to a plane; CD is I to the same plane.

Let A B, CD, meet the plane in B and D: join BD: :. A B, C D, B D, are in one plane. (Prop. VII.)

In the plane to which AB is -, draw de BD and = AB: join BE, A E, AD: then : AB is 1 to the plane, each of the 29 A B E, A B D, is a right Z; and : BD meets the parallels A B, CD; :: 2 ABD + 2 CDB = two right %s; but _ A BD is a right Z, :. LCDB is a right 2: ;.cdis + BD.

And :: AB = DE, and BD is common: and Z ABD=L ED B, :. AD= E B. Again, :: AB= DE, BE = A D, and A E is

:: _ ABE = Z EDA: but 2 A B E is a right 2,. EDA is so also, and E D is I DA: but Ed is + BD, .. to the plane through B D, da, but dc is in that plane; :. _ EDC is a right Z, or cd is 1 DE, but cd is 1 BD, :: to the plane through B D, D Е, i.e., to the same plane to which a B is 1.



Two straight lines, which are each of them parallel to the same

straight line, and not in the same plane with it, are parallel to one another.

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Let A B, CD, be each parallel to E F, and not in the same plane with it, A B shall be parallel to c D.

In EF take any pointa: from G, draw in the plane through EF, A B, GHI to EF: and in the plane through E F, CD, draw GK I EF.

Then :: EF is 1 to both gu and Gk, it is 1 to the plane HGK: and E r is parallel to AB: :. A B is I to H G K. Similarly cd is + to a GK, :. A B and cd being both + to the same plane are parallel. (Prop. VI.)

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If two straight lines meeting one another be parallel to two

others that meet one another, and are not in the same plane with the former two, the former two and the other two shall contain equal angles.



Let A B, BC, which meet in B, be respectively parallel to DE,

EF, meeting in E, but not in the same plane as A B, BC; the 2 ABC= _ DEP.

Take BA, B C, E D, DF, all =; join AD, C F, BE, A C, DF: then :: BA is and parallel to DE, 1. AD is = and parallel to BE: for the same reason CF is = and parallel to BE: :. AD and of being each of them = and parallel to BE, ::

they are = and parallel to each other: :: AC and Df, which join their extremities towards the same parts, are also = and parallel. But •: AB, BC = DE, EF, each to each, and AC = DF, .. _ ABC = DEF.


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To draw a straight line perpendicular to a plane from a given

point above it.

Let A be the given point, bu the given plane; it is required to draw from the point A, a straight line - to BH.

In the plane draw any straight line BC, and from a draw AD I to BC. Then if AD be also I to BH, AD is the I required: but if not, in by, draw DE BC: and from a draw AF I DE: AF is I to the plane BH.

Through r draw ou parallel to Bc.

Then : BC is - both to ED and Da, it is + to the plane through them; and an is parallel to BC.. Gu is also to the plane Eda and .. Go is to AF: and conversely af is -- to GH: and AF is 1 to DE. :. AF is + to the plane through ED and GH, i.e., to the plane Ba.

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CoR. Hence it follows that if af be a I to a plane, and PD a t to any line Bc in that plane, and AD be joined, that AD is Ito BC.


From a given point in a plane to draw a perpendicular to the


Let A be the given point in the plane: take any point B above the plane, from which draw BCI to the plane: and then

DI from a draw AD parallel to co: AD is the required. For BC being + to the plane, and AD parallel to BC, AD is also I to


the plane.


To find the perpendicular distance of two lines not in the same

plane. Let AB, CD, be the two lines. Through any point a in AB, draw ae || to cd, and let me be the plane passing through AB

and AE.

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From any point d in cp draw DF + to mn: and from fin MN, draw fg || to AE: from g draw gh || to FD: Gh is the perpendicular required.

For GF and cd being both ll to AE are || to each other; :: GF is in the same plane with cd and dy.. And thus GH, which is || to Fd, is in the same plane, and .. h is in cd.

Also, DF being 1 to the plane mn, gh which is || to it is also I MN: and .. to AB



and GF.

But gf is || to CD :: is also - to CD.
Also, gh is the shortest distance.

For if not, let ki be the shortest distance between AB and CD; then draw ko || to gh and FD. :. Ko is 1 to Mn; .. KI is > Ko, but Ko = GH, :. Ki is > GH, .. Gh is the shortest distance between AB and co.

In the last three problems, lines are assumed to be drawn from points || or 1 to other lines. This assumption, both in this and in other places, is necessary in order that the properties of figures may be exhibited; but the student must be reminded that he has as yet no practical method of drawing those lines, and that the effectual solution of these problems is one of the special objects of this work.


From the same point in a given plane there cannot be two straight

lines perpendicular to the plane, upon the same side of it; and there can be only one perpendicular to a plane from a point above the plane.

If possible, let AB, AC, be both + to a plane, from the same point A, and upon the same side of the plane.

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