Book XI. BC, DE, EF. But those planes to which the fame right. line is perpendicular, [by 14. 11.] are parallel: Therefore the planes paffing thro' A B, BC; DE, EF are par allel.

Therefore if two right lines touching one another be parallel to two right lines touching one another, but not in the fame plane; thofe planes which pass thro' them will be parallel. Which was to be demonftrated.

If any plane cuts two parallel planes, their common fections are alfo parallel.

For let fome plane EFGH cut the two parallel planes A B, CD, and let their common fections be EF, GH: I fay EF is parallel to GH.

K

For if they be not parallel, EF, GH being produced will either meet towards F, H, or towards E, G. Let them first be produced and meet towards F, H, viz. at K: Then because EFK is in the plane AB, all the points taken in EFK will be in the fame plane. But K is one of the points in E FK: Therefore K is in the plane A B: by the fame reafon K is also in the plane CD: Therefore the planes A B, CD being produced will meet. But they are supposed not to meet, because they are parallel: Therefore the right lines DEF, GH being produced will not meet towards F, H. In like manner we demonstrate that the right lines EF, G H being produced, will not meet towards E, G. But thofe lines which being both ways produced, do not meet, are [by 35. def. 1.] parallel: Therefore EF, G H are parallel. Therefore if any plane cuts two parallel planes; their common fections will be parallel. Which was to be demonftrated.

F

E

B

H

PROP.

PROP. XVII. THEOR.

If two right lines be cut by parallel planes, they will be cut in the fame ratio.

For let two right lines A B, CD, be cut [by the parallel planes GH, KL, MN in the points A, E, B, C, F, D: I fay, as the right line AE is to E B, fo is C F to F D.

For join AC, BD, A D, let AD meet the plane KL in the point x, and join EX, XF.

Then because the two parallel planes K L, M N are cut by the plane E BDX; [by 16. 11.] their common fections

ons AC, FX are parallel: But

because Ex is drawn parallel to one fide B D of the triangle A B D, [by 2. 6.] as A E is to EB, fo will Ax be to XD. Again, becaufe x F is drawn parallel to one fide A C of the triangle ADC; it will be as AX to X D, fo is cr to FD. But it has been proved that as A x is to X D, fo is AE to EB: Therefore [by 11. 5.] as A E is to E B, fo is

CF to F D.

Therefore if two right lines be cut by parallel planes, they will be cut in the fame ratio. Which was to be demonftrated.

PROP. XVIII. THE OR. "If a right line be at right angles to any plane, all the planes that pass thro' it will be at right angles to that plane.

For let the right line A B be at right angles to fome given plane: I fay all planes that pafs thro' AB will be at right angles to that given plane.

For

D

C

G A H

F B

E

Book XI: For let the plane DE pass thro' the right line AB, and let the right line CE be the common fection of the plane DE and the given plane: Take any point F in CE, and from the fame draw FG in the plane DE at right angles to CE. Then because the right line A B is perpendicular to the given plane, it will alfo [by 3. def. 11.] be perpendicular to all right lines lying in the given plane and touching it: Wherefore it is alfo perpendicular to CE; and fo A BF is a right angle. But GFB is also a right angle: Therefore [by 28. 1.] AB is parallel to F G. But A B is at right angles to the given plane: Wherefore [by 8. 11.] FG alfo will be at right angles to that fame plane. But [by 4. def. 11.] one plane is at right angles to another plane, when right lines. drawn at right angles to their common fection in one plane, are at right angles to the other plane: Therefore the right line F G drawn in one plane D E at right angles to the common fection CE is proved to be at right angles to the given plane. Confequently the plane D E is at right angles to the given plane. After the fame manner we demonftrate that every plane paffing thro' A B is perpendicular to the given plane.

If therefore a right line be at right angles to any given plane, all the planes that pass thro' it will be alfo at right angles to that given plane. Which was to be demonftrated.

PROP. XIX. THEOR.

If two planes cutting one another be at right angles to Some plane; their common fection will be at right angles to the fame plane.

For let two plane's A B, BC, cutting one another, be at right angles to fome given plane; and let their common fection be BD: I fay the right line B D is at right angles to that given plane.

For if it be not, [by 11. 1.] draw the right line DE from the point D in the plane A B at right angles to the right line A D, and in the plane BC draw D F at right an

gles

gles to CD. Then because the plane AB is at right angles to the given plane, and D E is drawn in the plane A B at right angles to their com

mon fection A D; DE will be perpendicular to the given plane. After the fame manner we demonftrate, that D F is also perpendicular to the given plane: Therefore there are two right lines drawn, on the fame fide, from the fame

point D at right angles to the

given plane, which [by 13.

II.] is impoffible. Therefore no right line but DB, the common fection of the planes AB, CD, can be drawn from the point D at right angles to the given plane.

Therefore if two planes cutting one another, be at right angles to fome plane; their common fection will also be at right angles to that fame plane. Which was to be demonftrated.

PROP. XX. THEOR.

If a folid angle be contained under three plane angles, any two of them taken together, are greater than the third.

For let a folid angle at A be contained under the three plane angles B AC, CAD, DAB: I fay any two of these angles BAC, CAD, DAB, taken together, are greater than the third.

For if the angles BAC, CAD, DAB be equal to one another, it is manifeft that any two of them, taken together, are greater than the third. If not, let B A C be the greateft of the three angles; and at the right line A B, and at the point a in it [by 23.1.] make the angle BAE in the plane paffing thro' B A C equal to the angle DAB; and [by 3. 1.] make A E equal to AD; alfo let the right line B E C drawn thro' E cut the right lines A B, A C in the points B, C, and join DB, D C..

B

E

Then

Book XI. Then because DA is equal to A E, and AB is common, the two fides DA, A B are equal to the two fides A E, AB, and the angle DAB is equal to the angle BAE: Therefore [by 4. 1.] the bafe DB is equal to the base B E. And because the two fides DB, DC are greater than BC, but D B has been proved to be equal to BE; the remainder DC will be greater than the remainder E c. And because D A is equal to A E, and AC is common, and the base DC is greater than the base EC; [by 25. 1.] the angle DAC will be greater than the angle E A C. But the angle DAB has been proved to be equal to the angle BAE; wherefore the angles DAB, DAC, taken together, are greater than the angle BA C. Also after the same manner we demonstrate, if any two other angles be taken, they are both together greater than the third angle remaining.

Therefore if a folid angle be contained under three plane angles, any two of them, taken together, are greater than the third. Which was to be demonstrated.

PROP. XXI.

THEOR.

Every folid angle is contained under plane angles that [taken together] are less than four right angles.

Let the folid angle at A be contained under the plane angles BA C, CAD, DAB: I fay the angles BA C, CAD, DAB, taken together, are less than four right angles.

For in each of the right lines A B, AC, AD, take any points B, C, D, and join B C, CD, DB. Then because the

B

folid angle at в is contained under the three plane angles CBA, ABD, CBD; any two of them, taken together [by 20. 11.] are greater than the third: Therefore the angles CBA, A B D are greater than the angle CBD. By the fame reason the angles BCA, ACD, taken together, are greater than the angle BCD; also the angles CDA, A D B are greater than the angle CDB: Wherefore the fix angles C B A, A B D, BCA, ACD, A DC, A D B are greater than the three angles CBD, BCD, CD B. But [by 32. 1.] the three angles CBD, BCD, CD B are equal to two right angles: Therefore the fix angles CBA, ABD, BCA, ACD, ADC, ADB are greater than two right angles. But becaufe the three

angles