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is in the plane, it can be produced in that plane : let it be produced to D; and let any plane pass through AD, and be turned about on AD, till it pass through C; and (I. def. 5.) because the points B, C are in this plane, BC is in it: therefore, there are two straight lines, ABC, ABD, in the same plane, that have a common segment AB, which (I. def. 3. cor.) is impossible. Therefore one part, &c.

PROP. II. THEOR. Two straight lines which cut one another, are in one plane: so also are three straight lines which meet one another, not in the same point.

Let two straight lines AB, CD, cut one another in E; AB, CD are in one plane: so also are three straight lines EC, CB, BE, which meet one another, not in the same point.

Let any plane pass through EB, produced, if necessary, and let it be turned about EB, till it pass through the point C. Then, (I. def. 5.) because the points E, C are in this plane, the straight line EC is in it. For the same reason BC is in the same; and (hyp.) EB is in it: therefore EC, CB, BE are in one plane. But C in the plane in which EC, EB are, in the same (XI. 1.) are CD, AB. Therefore AB, CD are in one plane. Wherefore two straight lines, &c.

PROP. III. THEOR. If two planes cut one another, their common section is a straight line.

Let two planes AB, BC cut one another, and let the line DB be their common section: DB is a straight line.

If it be not, from D to B, draw, in the plane AB, the straight line DEB, and in the plane BC the straight line DFB. Then two straight lines DEB, DFB enclose a space ; wbich (I. def. 3. cor.) is impossible : wherefore BD, the common section of the planes AB, BC, cannot but be a straight line.* Therefore, if two planes, &c.

PROP. IV. THEOR. If a straight line be perpendicular to each of two straight lines at their point of intersection, it is also perpendicular to the plane in which they are. Let the straight line EF be perpendicular to each of the straight

Thus, when the walls, floor, and ceiling of an apartment are planes, the lines in which they cut each other are straight lines. So likewise are the ridge and the eaves of a house, when the walls and the sloping surfaces of the roof are planes.

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lines AB, CD at their intersection E: EF is also perpendicular to the plane passing through AB, CD.

Take the straight lines EB, EC equal to one another, and join BC: in BC and EF take any points G and F, and join EG, FB, FG, FC. Then, in the triangles BEF, CEF, BE is equal to CE; EF common; and the angles BEF, CEF are equal, being (hyp.) right angles: therefore (I. 4.) BF is equal to CF. The triangle BFC is therefore isosceles : and (II. 5. cor. 5.) the square of BF is equal to the square of FG and the rectangle BG.GC. For the same reason, because (const.) the triangle BEC is isosceles, the square of BE is equal to the square GE and the rectangle BG.GC. To each of these add the square of EF; then the squares of BE, EF are equal to the squares of GE, EF and the rectangle BG.GC. But (I. 47.) the squares of BE, EF are equal to the square of BF, because BEF is a right angle ; and it has been shown, that the square of BF is equal to the square of FG and the rectangle BG.GC: therefore the square of FG and the rectangle BG.GC are equal to the squares of GE, EF and the rectangle BG.GC. Take the rectangle BG.GC from each, and there remains the square of FG, equal to the squares of GE, EF: wherefore (I. 48.) FEG is a right angle. In the same manner it would be proved, that EF is perpendicular to any other straight line drawn through E in the plane passing through AB, CD. But (XI. def. 1.) a straight line is perpendicular to a plane, when it makes right angles with all straight lines meeting it in that plane : therefore EF is perpendicular to the plane of AB, CD. Wherefore, if a straight line, &c.

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PROP. V. THEOR.

IF three straight lines meet all in one point, and a straight line be perpendicular to each of them at that point ; the three straight lines are in the same plane.

Let the straight line AB be perpendicular to each of the straight lines BC, BD, BE, at B, the point in which they meet : BC, BD, BE are in one and the same plane.

If not, let, if possible, BD and BE be in one plane, and BC be above it; and let a plane pass through AB, BC, the common section of which with the plane in which BD and BE are, is (XI. 3.) a straight line ; let this be BF. Therefore the three straight lines BA, BC, BF are all in one plane, viz., that which passes through AB, BC; and (XI. 4.) because AB is at right angles to each of the straight lines BD, BE, it is also at right angles to every straight line meeting it in that plane ; therefore ABF is a right angle: but

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(hyp.) ABC is also a right angle; therefore the angles ABF, ABC, in the same plane, are equal, which (I. ax. 9.) is impossible. Therefore BC is not above the plane of BD and BE: wherefore BC, BD, BE are all in the same plane. Therefore, if three straight lines, &c.

PROP. VI. THEOR.

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Ir two straight lines be at right angles to the same plane, they are parallel to one another.

Let the straight lines AB, CD be at right angles to the same plane BDE; AB is parallel to CD.

Let them meet the plane in the points B, D; join BD, and draw DE perpendicular to BD, in the plane BDĒ; make DE equal to AB, and join BE, AE, AD. Then, because AB is per. pendicular to the plane, each of the angles ABD, ABE is (XI. def. 1.) a right angle. For the same reason, CDB, CDE are right angles. And because AB is equal to DE, BD common, and the angle ABD equal to BDE, AD is equal (I. 4.) to BE. Again, in the triangles ABE, ADE, AB is equal to DE, BE to AD, and AE common ; therefore (1. 8.) the angle ABE is equal to EDA: but ABE is a right angle; therefore EDA is also a right angle, and ED perpendicular to DA: it is also perpendicular to each of the two BD, DC: therefore (XI. 5.) these three straight lines DA, DB, DC, are all in the same plane. But (XI. 2.) AB is in the plane in which are BD, DA : therefore AB, BD, DC are in one plane. Now (hyp.) each of the angles ABD, BDC is a right angle ; therefore (I. 28.) AB is parallel to CD. Wherefore, if two straight lines, &c.

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PROP. VII. THEOR. 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 AB, CD be parallel straight lines : the straight line which joins any points E and F in those lines, is in the same plane with the parallels.

If not, let it be, if possible, above the plane, as EGF; and in the plane ABDC, in which the parallels are, draw (I. post. 2.) the straight line EHF from E to F; and since EGF also is a straight line, the two straight lines EHF, EGF include a space between them, which (I. def. 3. cor.) is impossible. Therefore the straight line joining E, F, is not without the plane in which

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the parallels AB, CD are: it is therefore in that plane. Wherefore, if two straight lines, &c.

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PROP. VIII. THEOR. If one of two parallel straight lines be perpendicular to a plane, the other is also perpendicular to it.

Let AB, CD be parallel, and let one of them AB be perpendicular to the plane BDE; the other is also perpendicular to BDE.

Let AB, CD meet the plane in the points B, D, and join BD: then (XI. 7.) AB, CD, BD are in one plane. In the plane BDE draw DE perpendicular to BD, and equal to AB, and join BE, AE, AD. " Then, because AB is perpendicular to the plane, the angles ABD, ABE are (XI. def. 1.) right angles. Again, the angles ABD, CDB are together equal (I. 29.) to two right angles : but ABD is a right angle; therefore also CDB is a right angle, and CD perpendicular to BD. Then it would be proved, as in the sixth proposition of this book, that ED is perpendicular to DA: and being also (const.) perpendicular to BD, it is perpendicular (XI. 4.) to the plane passing through DA, DB. But (XI. 2.) A B is likewise in this plane; and so also is CD: DE, therefore, (XI. def. 1.) is perpendicular to CD. Therefore, since CD is perpendicular to each of the two straight lines DB, DE, it is perpendicular (XI. 4.) to the plane BDE in which those lines are. Wherefore, if one, &c.

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PROP. IX. THEOR.

Two straight lines which are each of them parallel to the same straight line, and are not both in the same plane with it, are parallel to one another.

Let AB, CD be each of them parallel to EF, and not both in the same plane with it; AB is parallel to CD.

From any point G in EF, draw GH, in the plane passing through EF, AB, at right angles to EF; and in the plane passing through EF, CD, draw GK also at right angles to EF. Then, because EF is perpendicular both to GH and GK, it is perpendicular (XI. 4.) to the plane HGK passing through them. But EF is parallel to AB; therefore AB is at right angles (XI. 8.) to HGK. For the same reason, CD is likewise at right angles to HGK. Therefore (XI. 6.) since AB, CD are each of them at right angles to HGK, AB is parallel to CD. Wherefore two straight lines, &c.

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Schol. The same has been proved (I. 30.) respecting straight lines in the same plane: therefore, universally, straight lines which are parallel to the same straight line, are parallel to one another.

PROP. X. THEOR.

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 first two; the first two and the other two contain equal angles. *

Let the straight lines AB, BC, which meet one another, be parallel to DE, EF, which also meet one another, but are not in the same plane with AB, BC: the angle ABC is equal to DEF.

Take BA, BC, ED, EF all equal to one another; and join AD, CF, BE, AC, DF. Because BA is equal and parallel to ED, therefore AD is (I. 33.) both equal and parallel to BE. For the same reason, CF is equal and parallel to BE. Therefore AD and CF being each of them parallel to BE, are (XI. 9.) parallel to one another. They are also (I. ax. 1.) equal ; and AC, DF join them towards the same parts; and therefore (I. 33.) AC is equal and parallel to DF. And because AB, BC are equal to DE, EF, and AC to DF; the angle ABC is equal (I. 8.) to DEF. Therefore, if two straight lines, &c.

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PROP. XI. PROB. To draw a straight line perpendicular to a plane, from a given point above it.

Let A be the given point above the plane BH; it is required to draw from A a perpendicular to BH.

In the plane draw any straight line BC, and (I. 12.) from A draw AD perpendicular to BC. Then, if AD be also perpendicular to the plane BH, the thing required is done. But if it be not, from D (I. 11.) draw DE, in the plane BH, at right angles to BC; from A draw AF perpendicular to DE; and through F draw (1. 31.) GH parallel to BC. Then, because BC is at right angles to ED and DA, BC is at right angles (XI. 4.) to the plane passing through ED, DA: and GH

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* Or supplemental ones ; as will be plain after the demonstration here given, if AB be produced through B. This generalizes the 4th corollary to the 34th proposition of the first book.

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