PROP XXXVIII. THEOREM. “ If a plane be perpendicular to another plane, and a straight line be « drawn from a point in one of the planes perpendicular to the other plane, this straight line shall fall on the common section of the planes. “Let the plane CD be perpendicu- “plane, meets it; therefore FGE is " lar to the plane AB, and let AD be a right angle (3. def. 11): But EF “ their common section ; if any point“ is also at right angles to the plane “E be taken in the plane CD, the per AB; and therefore EFG is a right “pendicular drawn from É to the angle : Wherefore two of the angles plane AB shall fall on AD. “ of the triangle EFG are equal toge" For, if it does not, let it, if possi- “ther to two right angles; which is “ble, fall elsewhere, as EF; and let « absurd: There E " it meet the plane AB in the point “fore the perpen“F; and from F draw, (12. 1.) in the “ dicular from the А plane AB a perpendicular FG to “point E to the " DA, which is also perpendicular (4. “ plane AB does “ def. 11.) to the plane CD; and join “not fall else -В - EG: Then because FG is perpen “ where than upon the straight line “ dicular to the plane CD, and the “AD; it therefore, falls upon it. If “ straight line EG, which is in that “ therefore a plane," &c. Q.E.D. с PROP. XXXIX. THEOREM. 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 opposite planes therefore, because KL, BA are each CF, AH of the solid parallclopiped of them parallel to DC, and not in the D K same plane with it, KL is parallel (9. 11.) to Bd: And because KL.MN E 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, N XR; and DG the diameter of the so lid parallelopiped AF: YS and DG AF, be divided each into two equal do meet, and cut one another into two parts in the points K, L, M, N; X, equal parts. O, P,R; and join KL, MN, XO, PR: Join DY, YE,BS,SG. Because DX And because DK, CL are equal and is parallel to Oé, the alternate angles parallel, KL is parallel (33. 1.) to DC: DŠY, YOE are equal (29. 1.) to one For the same reason, ÀN is parallel another: And because DX is equal to to BA; And BA is parallel to DC; OE, and XY to YO, and contain e qual angles, the base DY is equal (4. (33. 1.) to EG: And DE, BG join 1.) to the base YE, and the other an their extremities; therefore DE is e. qual and parallel (33. 1.) to BG: And ÕG, YS are drawn from points in the one to points in the other; and are E therefore in one plane: Whence it is manifest, that DĠ, YS must meet one IT another; let them meet in T: And because DE is parallel to BG, the al ternate angles EDT, BGT are equal ; B (29. 1.) and the angle DTY, is equal (15. 1.) to the angle GTS: Therefore N in the triangles DTY, GTS there are two angles in the one equal to two gles are equal; therefore the angle angles in the other, and one side equal XYD is equal to the angle OYE, and to one side, opposite to two of the DYE is a straight (14. 1.) line: For equal angles, viz. DY to GS; for they the same reason BSG is a straight are the halves of DE, BG : Therefore line, and BS equal to SG: And be the remaining sides are equal, (26. 1.) cause CA is equal and parallel to DB, each to each. Wherefore DT is equal and also equal and parallel to EG; to TG, and YT equal to TS. Where. therefore DB is equal and parallel fore, if in a solid, &c. Q. E. D. PROP. XL. THEOREM. If there be two triangular prisms 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, GHKL- the triangle GHK, the prism ABCDEF MN be of the same altitude, the first is equal to the prism GHKLMN. whereof is contained by the two tri Complete the solids AX, GO; and angles ABE, CDF, and the three pa- because the parallelogram AF is rallelograms AD, DE, EC; and the double of the triangle GHK; and the other by the two triangles GHK, parallelogram HK double (34. 1.) of LMN, and the three parallelograms the same triangle; therefore the paLH, HN, NG; and let one of them rallelogram AF is equal to HK. But have a parallelogram AF, and the solid parallelopipeds upon equal bases, other a triangle GHK for its base; and of the same altitude, are equal if the parallelogram AF be double of (31. 11.) to one another. Therefore the solid AX is equal to the solid M and the prism ABCDEF is half X (28. 11.) of the solid GO. Therefore N the prism ABCDEF is equal to the HI prism GHKLMN. Wherefore, if there be two, &c. Q. E, D. GO; 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 ils 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. more Let AB and C be two unequal are in DE: And let the divisions in magnitudes, of which AB is the great AB be AK, KH, HB; and the divier. If from AB there be taken more sions in Ed be EF, FG, GE: And than its half, and from D because DE is greater than AB, and the remainder that EG taken from DE is not greater than its half, and so on; K than its half, but BH taken from AB there shall at length re is greater than its half; therefore the main a magnitude less H rernainder GD is greater than the rethan C. mainder HA. Again, because GD is For C may be multi greater than HA, and that GF is not plied so as at length to B C! H greater than the half of GD, but HK become greater than is greater than the half of Ha; thereAB. Let it be so multiplied, and let fore the remainder FD is greater than DE its multiple be greater than AB, the remainder AK. And FD is equal and let DE be divided into DF, FG, to C, therefore C is greater than AK; GE, each equal to C. From AB take that is, AK less than c. Q. E. D. BH greater than its half, and from And if only the halves be taken the remainder AH take HK greater away, the same thing may in the same than its half, and so on, until there way be demonstrated. be as many divisions in AB as there PROP. I. THEOREM. Similar polygons inscribed in circles are to one another as the squares of their diamelcrs. Let ABCDE, FGHKL be two cir- polygons are divided into similar tricles, and in them the similar poly, angles; the triangle ABE, FGL, are gons ABCDE, FGHKL; and let BM, GN, be the diameters of the circies : As the square of BM is to the square of GN, so is the polygon ABCDE to B L the polygon FGHKL. Join BE, AM, GL, FN: And be M cause the polygon ABCDE is similar to the polygon FGHKL, and similar similar and equiangular (6.6.); and of GN, is the duplicate (20. 6.) ratio therefore the angle A EB is equal to of that which BM has to GN; and the angle FLG: But AEB is equal the ratio of the polygon ABCDE to (21.3.) to AMB, because they stand the polygon FGHKL is the duplicate upon the same circumference; and the angle FLG is, for the same reason, equal to the angle FNG: There B. E G fore also the angle AMB is equal to L FNG: And the right angle BAM is equal to the right (31. 3.) angle GFN; M N wherefore the remaining angles in the H K triangles ABM, FGN are equal, and they are equiangular to one another : (20. 6.) of that which BA has to CF: Therefore as BM to GN, so (4.6.) Therefore as the square of BM to the is BA to GF; and therefore the du- square of GN, so is the polygon plicate ratio of BM to GN is the same ABCDE to the polygon FGHKL. (10. def. 5. & 22. 5.) with the dupli- Wherefore similar polygons, &c. cate ratio of BA to ĠF: But the ra. Q. E. D. tio of the square of BM to the sqnare PROP. II. THEOREM. Circles are to one another as the squares of their diameters. Let ABCD, EFGH be two circles, the square EFGH is greater than and BD, FH their diameters: As the half of the circle. Divide the circumsquare of BD to the square of FH, so ferences EF, FG, GH, HE, each is the circle ABCD, to the circle into two equal parts in the points K, EFGH. L, M, N, and join EK, KF, FL, LG, For, if it be not so, the square of GM, MH, HN, NE: Therefore each BD shall be to the square of FH, as of the triangles EKF, FLG, GMH, the circle ABCD is to some space HNE is greater than half of the segeither less than the circle EFGH, or ment of the circle it stands in ; begreater than it.* First, let it be to a cause, if straight lines touching the space S less than the circle EFGH; circle be drawn through the points and in the circle EFGH describe the K, L, M, N, and parallelograms upsquare EFGH: This square is greater on the straight lines EF, FG, GH, than half of the circle EFGH; be- HE, be completed; each of the tricause if, through the points E, F, G, angles EKF, FLG,GMH, HNE shall H, there be drawn tangents to the be the half (41. 1.) of the parallelocircle, the square EFGH is half (41. gram in which it is : But every seg1.) of the square described about the ment is less than the parallelogram circle; and the circle is less than the in which it is : Wherefore each square described about it; therefore of the triangles EKF, FLG, GMH, For there is some square equal to the circle ABCD; let P be the side of it : and to three straight lines BD, FH, and P, there can be a fourth proportional ; let this be Q: Therefore the squares of these four straight lines are proportionals; that is, to the squares of BD, FH, and the circle ABCD, it is possible there may be a fourth proportional. Let this be S. And in like manner are to be understood some things in some of the following propositions. E N HNE is greater than half the segment greater (14. 5.) than the polygon of the circle which contains it: And EKFLGMHN: But it is likewise if these circumferences before-named less, as has been demonstrated ; be divided each into two equal parts, which is impossible. Therefore the and their extremities be joined by square of BD is not to the square of straight lines, by continuing to do FH, as the circle ABCD is to any space less than the circle EFGH. In R E the same manner, it may be demonN strated, that neither is the square of B в D s FH to the square of BD, as the cir. cle EFGH is to any space less than PL the circle ABCD. Nor is the square G of BD to the square of FH, as the this, there will at length remain seg. circle ABCD is to any space greater ments of the circle, which, together, than the circle EFGH: For, if posshall be less than the excess of the sible, let it be so to T, a space greater circle EFGH, above the space S: than the circle EFGH: Therefore, Because, by the preceding Lemma, if inversely, as the square of FH to the from the greater of two unequal mag. square of BD, so is the space T to nitudes there be taken more than its the circle ABCD. But as the space half, and from the remainder more than its half, and so on, there shall Х R K к І H nitudes. Let then the segments EK, M KF, FL, LG, GM, MH, HN, NE be those that remain and are together less than the excess of the circle EFGH, above S: Therefore the rest of the circle, viz. the polygon T EKFLGMHN, is greater than the space S. Describe likewise in the cir. T is to the circle ABCD, so is the cle ABCD the polygon AXBOCPDR circle EFGH to some space, which similar to the polygon EKFLGMHN: must be less (14. 5.) than the circle As therefore, the square of BD is to ABCD, because the space T is greater, the square of FH, so (1. 12.) is the by hypothesis, than the circle EFGH. polygon AXBOCPDR to the poly. Therefore as the square of FH is to gon EKFLGMHN; But the square the square of BD, so is the circle of BD is also to the square of FH, as EFGH to a space less than the circle the circle ABCD is to the space S: ABCD, which has been demonstrated Therefore as the circle ABCD is to to be impossible : Therefore the the space S, so is (11. 5.) the poly, square of BD is not to the square of gon AXBOCPDR to the polygon FH as the circle ABCD is to any EKFLGMHN: But the circle ABCD space greater than the circle EFGH: is greater than the polygon contained And it has been demonstrated, that in it: wherefore the space S, is neither is the square of BD to the P I • For, as in the foregoing note at“, it was explained how it was possible there could be a fourth proportional to the squares of BD, FH, and the circle ABCD, which was named 'S. So, in like manner, there can be a fourth proportional to this other space, named T, and the circles ABCD, EFGH. And the like is to be understood in some of the following propositions. |