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shall be at right (1.7) angles to the same plane ; wherefore GH is at right angles to the plane through ED, DA, and is perpendicular (1. Def. 1) to every straight line meeting it in that plane. But AF, which is in the plane through ED, DA, meets it; therefore GH is perpendicular to AF; and consequently AF is perpendicular to GH; and AF is also perpendicular to DE; therefore AF is perpendicular to each of the straight lines GH, DE. But if a straight line stands at right angles to each of two straight lines in the point of their intersection, it shall also be at right angles to the plane passing through them; and the plane passing through ED, GH, is the plane BH; therefore AF is perpendicular to the plane BH; so that, from the given point A, above the plane BH, the straight line AF is drawn per: pendicular to that plane. CoR.-If it be required from a point C in a plane to

erect a perpendicular to that plane, take a point A above the plane, and draw A perpendicular to the plane; then, if from C a line be drawn parallel to AF, it will be the perpendicular required; for, being parallel to AF, it will be perpendicular to the same plane to which AF is perpendicular.

PROPOSITION XI. THEOREM. From the same point in a given plane, there cannot be two straight lines at right angles to the plane, upon the same side of it; and there can be but one perpendicular to a plane from a point above the plane.

For, if it be possible, let the two straight lines AC, AB, be at right angles to a given plane from the same point A in the plane, and upon the same side of it; and let a plane pass through BA, AC; the common section of this with the given plane is a straight line passing through A (1.3). Let DAE be their common section; therefore D A E the straight lines AB, AC, DAE, are in one plane; and because CA is at right angles to the given plane, it shall make right angles with every straight line meeting it in that plane. But DAE, which is in that plane, meets CA; therefore CAE is a right angle. For the same reason, BAE

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is a right angle; wherefore the angle CAE is equal to the angle BAE; and they are in one plane; which is impossible. Also, from a point above a plane, there can be but one per: pendicular to that plane; for if there could be two, they would be parallel (1.6) to one another; which is absurd.:

PROPOSITION XII. THEOREM. Planes to which the same straight line is perpendicular, are parallel to one another. - Let the straight line AB be perpendicular to each of the planes CD, EF; these planes are parallel to one another..

If not, they shall meet one another when produced ; let them meet; their common section shall be a straight line GH, in which take any point K, and join AK, BK; then, because AB is perpendicular to the plane EF, it is perpendicular (I. Def. 1) c A to the straight line BK which is in that plane; 4 therefore ABK is a right angle; for the same reason, BAK is a right angle; wherefore the VDEN two angles ABK, BAK, of the triangle ABK, are equal to two right angles; which is impossible (Pl. Ge. I. 17); therefore the planes CD, EF, though produced, do not meet one another; that is, they are parallel (I. Def. 6.)

PROPOSITION XIII. THEOREM. If two straight lines meeting one another be parallel to two straight lines which meet one another, but are not in the same plane with the first two, the plane which passes through these is parallel to the plane passing through the others. - Let AB, BC, two straight lines meeting one another,' be parallel to DE, EF, that meet one another, but are not in the same plane with AB, BC; the planes through AB, BC, and DE, EF, shall not meet, though produced.

From the point B draw BG perpendicular (I. 10) to the plane which passes through DE, EF, and let it meet that plane in G; and through Ğ draw GH paral- E -"} lel to ED, and GK parallel to EF. And ai i GN because BG is perpendicular to the plane 109YYK through DE, EF, it shall make right angles al with every straight line meeting it in that V DIN

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plane (I. Def. 1). But the straight lines GH, GK, in that plane meet it; therefore each of the angles BGH, BGK, is a right angle. And because BA is parallel (I. 8) to GH (for each of them is parallel to DE), the angles ĜBA, BGH, are together equal (Pl. Ge. I. 29) to two right angles. And BGH is a right angle; therefore also GBA is a right angle, and GB perpendicular to BA. For the same reason, GB is perpendicular to BC. Since therefore the straight line GB stands at right angles to the two straight lines BA, BC, that cut one another in B, GB is perpendicular (I. 4) to the plane through BA, BC; and it is perpendicular to the plane through DE, EF; therefore BG is perpendicular to each of the planes through AB, BC, and DE, EF. But planes to which the same straight line is perpendicular, are parallel (I. 12) to one another. Therefore the plane through AB, BC, is parallel to the plane through DE, EF.

PROPOSITION XIV. THEOREM. If two parallel planes be cut by another plane, their common sections with it are parallels. . Let the parallel planes AB, CD, be cut by the plane EFHG, and let their common sections. with it be EF, GH; EF is parallel to GH.

For the straight lines EF and GH are in the same plane, namely, EFHG, which cuts the planes N _ AB and CD; and they do not meet though produced; for the planes in which they are, do not meet; therefore EF and A GH are parallel.

PROPOSITION XV. THEOREM. If two parallel planes be cut by a third plane, they have the same inclination to that plane.

Let AB and CD be two parallel planes, and EH a third plane cutting them. The planes AB and CD are equally inclined to EH.

Let the straight lines EF and GH be the common sections of the plane EH with the two planes AB and CD; and from K any point in EF, draw in the plane EH the straight line KM at right angles to EF, and let it meet GH in

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L; draw also KN at right angles to EF in the plane AB; and through the straight lines KM, KN, let a plane be made to pass cutting the plane CD in the line LO. And because EF and GH are the com- BLU mon sections of the plane EH with the two parallel planes AB and CD, EF is parallel to GH (I. 14). But EF is at right angles to the plane that passes through KN and KM (1.4), because it is at right angles to the lines KM and KN; therefore GH is also at right angles to the same plane (I. 7), and it is therefore at right angles to the lines LM, LO, which it meets in that plane. Therefore, since LM and LO are at right angles to LG, the common section of the two planes CD and EH, the angle OLM is the inclination of the plane CD to the plane EH (I. Def, 4). For the same reason the angle MKN is the inclination of the plane AB to the plane EH. But because KN and LO are parallel, being the common sections of the parallel planes AB and CD with a third plane, the interior angle NKM is equal to the exterior angle OLM; that is, the inclination of the plane AB to the plane EH, is equal to the inclination of the plane CD to the same plane EH.

- PROPOSITION XVI. THEOREM. If two straight lines be cut by parallel planes, they shall be cut in the same ratio.

Let the straight lines AB, CD, be cut by the parallel planes GH, KL, MN, in the points A, E, B; C, F, D. As AE is to EB, so is CF to FD.

Join AC, BD, AD, and let AD meet the plane KL in the point X; and join EX, XF. Because the two parallel planes KL, MN, are cut by the plane EBDX, G the common sections EX, BD, are parallel LR Ja (I. 3). For the same reason, because the two parallel planes GH, KL, are cut by the plane AXFC, the common sections AČ, XF, are parallel. And because EX is parallel to BD, m/ å side of the triangle ABD, as AE to EB, so B is (Pl. Ge. VI. 2) AX to XD. Again, because XF is parallel to AC, a side of the triangle ADC, as AX to XD, so

is CF to FD. And it was proved that AX is to XD, as AE to EB. Therefore (Pl. Ge. V. 1l), as AE to EB, so is CF to FD.

PROPOSITION XVII. THEOREM. If a straight line be at right angles to a plane, every plane which passes through it shall be at right angles to that plane.

Let the straight line AB be at right angles to a plane CK; every plane which passes through AB shall be at right angles to the plane CK.

Let any plane DE pass through AB, and let CE be the common section of the planes DE, CK; take any point F in CE, from which draw FG in the plane D G A II DE at right angles to CE. And because AB is perpendicular to the plane CK, therefore it is also perpendicular to every straight line meeting it in that plane (). C F B I Def. 1); and consequently it is perpendicular to CE. Wherefore ABF is a right angle; but GFB is likewise a right angle; therefore AB is parallel to FG. And AB is at right angles to the plane CK; therefore FG is also at right angles to the same plane (I. 7). But one plane is at right angles to another plane when the straight lines drawn in one of the planes, at right angles to their common section, are also at right angles to the other plane (I. Def. 2); and any straight line FG in the plane DE, which is at right angles to CE, the common section of the planes, has been proved to be perpendicular to the other plane CK; therefore the plane DE is at right angles to the plane CK. In like manner, it may be proved that all the planes which pass through AB are at right angles to the plane CK.

PROPOSITION XVIII. THEOREM. If two planes cutting one another be each of them perpendicular to a third plane, their common section shall be perpendicular to the same plane.

Let the two planes AB, BC, be each of them perpendicular to a third plane, and let BD be the common section of the first two; BD is perpendicular to the plane ADC.

From D in the plane ADC, draw DE perpendicular to

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