* For if it does not, let it, if poflible, fall elsewhere, as EF; and Book XI. * let it meet the plane AB in the point F; and from F draw", in " the plane AB, a perpendicular FG to DA, which is also per- a. 11. 1. “ pendicular b to the plane CD; and join EG. then because FG is b.4.Dcf.11. “ perpendicular to the plane CD, C, “ and the straight line EG, which " is in that plane, meets it ; there“ fore FGE is a right angle c. but “ EF is also at right angles to the “ plane AB; and therefore EFG “is a right angle. wherefore two

B “ of the angles of the triangle “ EFG are equal together to two right angles; which is absurd. " therefore the perpendicular from the point E to the plane AB " does not fall elfewhere than upon the straight line AD. it there“ fore falls upon it. If therefore a plane, &c. Q. L. D.”

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N a solid parallelepiped, if the sides of two of the op- See N.

posite planes be divided each into two equal parts, the common section of the planes passing through the points of division, and the diameter of the folid parallelepiped cut cach other into two equal parts.

Let the sides of the opposite planes D K F CF, AH of the folid

Y parallelepiped AF, be divided each into


E two equal parts in the points K, L, M, N; X, O, P, R; and join KL, MN, XO, PR. and because DK, CL are equal B and parallel, KL is parallel to DC. for P

the same reason,
MN is parallel to



2. 33. 1.

c. 29. 1.

d. 41.

Book XI. BA. and BA is parallel to DC; therefore because KL, BA are

each of them parallel to DC, and not in the fame plane with it, KL b. 9. II.

is parallel to BA. and because KL, MN are each of them pa-
rallel to BA, and not in the fame plane with it, KL is parallel b
to MN; wherefore KL, MN are in one plane. In like manner,
it may be proved that XO, PR are in one planc. Let YS be the
common section of the planes KN, XR; and DG the diameter of
the solid parallelepiped AF. YS and DG do meet, and cut one
another into two equal parts.

Join DY, YE, BS, SG. because DX is parallel to OE, the al-
ternate angles DXY, YOE are equal to one another, and because
DX is equal to OE,
and XY to YO,

, D K
and contain equal

angles, the base DY
is equal to the

base YE, and the
other angles are e-
qual; therefore the

angle XYD is e-
qual to the angle
OYE, and DYE is

e. 14. 1. a straight line. for
the same reason P

BSG is a straight
line, and BS equal


to SG. and because
CA is equal and parallel to DB, and also equal and parallel to

EG; therefore DB is equal and parallel 6 to EG. and DE, BG 2. 33. 1. join their extremities; therefore DE is equal and parallelo to BG.

and DG, YS are drawn from points in the one to points in the
other, and are therefore in one plane. whence it is manifest that
DG, YS must meet one another ; let them meet in T. and be.

cause DE is parallel to BG, the alternate angles EDT, BGT are f. 15. 1. equal“; and the angle DTY is equal f to the angle GTS. therefore

in the triangles DTY, GTS there are two angles in the one equal
to two angles in the other, and one side equal to one side, oppo-

site to two of the equal angles, viz. DY to GS; for they are the S: 26.1. halves of DE, BG. thesefore the remaining fides are equals, each

to each. wherefore DT is equal to TG, and YT equal to TS.
Therefore if in a solid, &c. Q. E. D.

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Book XI.

IF there be two triangular prisms of the fame altitude,

the base of one of which is a parallelogram, and the base of the other a triangle ; if the parallelogram be double of the triangle, the prisins shall be equal to one another.

Let the prisms ABCDEF, GHKLMN be of the same altitude, the first whereof is contained by the two triangles ABE, CDF, and the three parallelograms AD, DE, EC; and the other by the two triangles GHK, LMN and the three parallelograms LH, HN, NG; and let one of them have a parallelogram AF, and the other a triangle GHK for its bafe ; if the parallelogram AF be double of the triangle GHK, the prism ABCDEF is equal to the prism GHKLMN.

Complete the folids AX, GO, and because the parallelogram AF is double of the triangle GAK; and the parallelogram HK double · of the faine triangle ; therefore the parallelogram AF is a. 34. 8.

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equal to HK. but folid parallelepipeds upon equal bases, and of the same altitude are equal 6 to one another. therefore the folid b. 31. 31, AX is equal to the folid GO. and the prism ABCDEF is halfic c. 23.11. of the solid AX; and the prism GHKLMN half of the solid GO. therefore the prilin ABCDEF is equal to the prism GHKLMN. Wherefore if there be two, &c. Q. E. D.'

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Which is the first Proposition of the tenth Book, and is necessary

to some of the Propositions of this Book. F from the greater of two unequal magnitudes there

be taken more than its half, and from the remainder more than its half; and so on. there shall at length remain a magnitude less than the least of the proposed magnitudes.


Let AB and C be two unequal magnitudes, of which AB is the greater. if from AB there be taken more than its half, and from the remainder more A

than its half, and so on; there Mall at length
remain a magnitude less than C.

For C may be multiplied so as at length to

become greater than AB. let it be fo multi-
plied, and let DE its multiple be greater than H
AB, and let DE be divided into DF, FG,

G GE, cach equal to C. from AB take BH greater than its half, and from the remainder Al take TIK greater than its half, and so on until there be as many divisions in AB as there B C E are in DE. and let the divisions AK, KH, HB be as many as the divisions DF, FG, GE. and because DE is greater than AB, and that EG taken from DE is lefs than its

hulf, but BH taken from AB is greater than its half, therefore the Book XII. reman ler GD is greater than the remainder HA. again, because G” is greater than HA, and that GF is the half of GD, but HK is greater than the half of HA ; therefore the remainder FD is greater than the remainder AK. and FD is equal to C, therefore C is greater than AK ; that is, AK is less than C. Q. E. D.

And if only the halves be taken away, the same thing may in the faine way be demonstrated,


IMILAR polygons inscribed in circles, are to one

another as the squares of their diameters.

Let ABCDE, FGHKL be two circles, and in them the fimilar polygons ABCDE, FGHKL ; and let BM, GN be the diameters of the circles. as the square of BM is to the fquare of GN, so is the polygon ABCDE to the polygon FGHKL.

Join LE, AM, GL, FN. and because the polygon ABCDE is fimilar to the polygon FGHKL, the angle BAE is equal' to the angle a. 1. Del.6, GFL, and as BA 10 AE, fo . is GF to FL. wherefore the two triangles BAE, GFL having one angle in one equal to one angle in the

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other, and the sides about the equal angles proportionals, are equiangular"; and therefore the angle AEB is equal to the angle FLG. b. 6. 6, but AEB is equal to the angle AMB, because they stand upon the c. 21. 3. fame circumference ; and the angle FLG is, for the same reason, equal to the angle FNG. therefore also the angle AMB is equal to FNG. and the right angle BAM is equal to the right d angle GFN; d. 31 3, wherefore the remaining angles in the triangles ABM, TGN are equal, and they are equiangular to one another, therefore as BM to

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