E

"For, if it does not, let it, if possible, fall elsewhere, as EF; and let it meet the plane AB in the point F; and from F draw (12. 1.) in the plane AB a perpendicular FG to DA, which is also perpendicular (4. def. 11.) to the plane CD; and join EG. then because FG is perpendicular to the plane CD, and the straight line C EG, which is in that plane, meets it; therefore FGE is a right angle (3. def. 11.): but EF is also at right angles to the plane AB; and therefore A EFG is a right angle: wherefore two of the angles of the triangle EFG are equal together to two right

D

B

angles; which is absurd: therefore the perpendicular from the point E to the plane AB, does not fall elsewhere than upon the straight line AD; it therefore falls upon it. If therefore a plane," &c. Q. E. D.

PROP. XXXIX. THEOR.

In a solid parallelopiped, if the sides of two of the opposite 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 solid parallelopiped cut each other into two equal parts.*

Let the sides of the D opposite planes CF, AH of the solid parallelopiped AF, be divided each into two equal parts in the points K, L, M, N; X, O, P, R; and join KL, MN, XO, PŘ: and because DK, CL are equal and parallel, KL is parallel (33. 1.) to DC: for B the same reason, MN is parallel to BA: and BA is parallel to DC; therefore because KL, BA, are each of them

C

*

L

T

E

parallel to DC, and not in the same plane with it, KL is parallel (9. 11.) to BA: and because KL, MN are each of them parallel to BA, and not in the same plane with it, KL is parallel (9. 11.) to MN; wherefore KL, MN are in one plane. In like manner, it may be proved, that XO, PR are in one plane. Let YS be the common section of the planes KN, XR; and DG the diameter of the solid parallelopiped 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 alternate angles DXY, YOE are equal (29. 1.) to one another: and because DX is D

F

equal to OE, and

XY to YO, and contain equal angles, the base DY is equal (4. 1.) to the base YE, and the other angles are equal; therefore the angle XYD is equal to the angle OYE and DYE is a straight (14. 1.) line: for the same BSG is a straight line, and BS equal to SG: and because CA is equal and parallel to

reason

B

K

DB, and also equal and parallel to EG, therefore DB is equal and parallel (9.11.) to EG: and DE, BG join their extremities, therefore DE is equal and parallel (33. 1.) 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 because DE is parallel to BG, the alternate angles EDT, BGT are equal (29. 1.); and the angle DTY is equal (15. 1.) 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, opposite to two of the equal angles, viz. DY to GS; for they are the halves of DE, BG: therefore the remaining sides are equal (26. 1.) each to each. Wherefore DT is equal to TG, and YT equal to TS. Wherefore, if in a solid, &c. Q. E. D.

PROP. XL. THEOR.

If there be two triangular prisnis of the same 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 prisms 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 base; if the parallelogram AF be double of the triangle GHK, the prism ABCDEF is equal to the prism GHKLMN.

Complete the solids AX, GO; and because the parallelogram AF is double of the triangle GHK; and the parallelogram HK

double (34.1.) of the same triangle; therefore the parallelogram AF is equal to HK. But solid parallelopipeds upon equal bases, and of the same altitude, are equal (31. 11.) to one another. Therefore the solid AX is equal to the solid GO; and the prism ABCDEF is half (28. 11.) of the solid AX; and the prism GHKLMN half (28. 11.) of the solid GO. Therefore the prism ABCDEF is equal to the prism GHKLMN. Wherefore, if there be two, &c. Q. E. D.

THE

ELEMENTS OF EUCLID.

BOOK XII.

LEMMA I.

Which is the first proposition of the tenth book, and is necessary to some of the propositions of this book.

IF 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.*

A

D

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

K

For C may be multiplied so, as at length to become greater than AB. Let it be so multiplied, and let DE multiple be greater than AB, and let DE be divided into DF, FG, GE, each equal to C. From AB take BH greater than its half, and from H the remainder AH take HK greater than its half, and so on, until there be as many divisions in AB as there are in DE: and let the divisions in AB be AK, KH, HB; and the divisions in ED be DF, FG, GE. And because DE is greater than AB, and

* See Note.

F.

G

BCE

1

that EG taken from DE is not greater than its half, but BH taken from AB is greater than its half; therefore the remainder GD is greater than the remainder HA. Again, because GD is greater than HA, and that GF is not greater than 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 same way be demonstrated.

PROP. I. THEOR.

SIMILAR polygons inscribed in circles are to one another as the squares of their diameters.

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

Join BE, AM, GL, FN: and because the polygon ABCDE is similar to the polygon FGHKL, and similar polygons are divided into similar triangles; the triangles ABE, FGL are similar and

equiangular (6. 6.); and therefore the angle AEB is equal to the angle FLG: but AEB is equal (21. 3.) to AMB, because they stand upon the same 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 (31. 3.) angle GFN; wherefore the remaining angles in the triangles ABM, FGN are equal, and they are