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BOOK XII. one another: Therefore as BM to GN, so is BA to GF; and therefore the duplicate ratio of BM to GN, is the € 4. 6. same with the duplicate ratio of BA to GF: But the ra& 22. 5. tio of the square of BM to the square of GN, is the dupli20. 6. cate ratio of that which BM has to GN; and the ratio of

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the polygon ABCDE to the polygon FGHKL is the duplicates of that which BA has to GF: Therefore as the square of BM to the square of GN, so is the polygon ABCDE to the polygon FGHKL. Wherefore similar polygons, &c. Q.E.D.

PROP. II. THEOR.

See N. CIRCLES are to one another as the squares of their diameters.

Let ABCD, EFGH be two circles, and BD, FH their diameters: As the square of BD to the square of FH, so is the circle ABCD, to the circle EFGH.

For, if it be not so, the square of BD shall be to the square of FH, as the circle ABCD is to some space either less than the circle EFGH, or greater than it*. First let it be to a space S less than the circle EFGH; and in the circle EFGH describe the square EFGH. This square is greater than half of the circle EFGH; because if, through the points E, F, G, H, there be drawn tangents 47. 1. to the circle, the square EFGH is half of the square de

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.

scribed about the circle: and the circle is less than the BOOK XII. square described about it; therefore the square EFGH is greater than half of the circle. Divide the circumferences EF, FG, GH, HE, each into two equal parts in the points K, L, M, N, and join EK, KF, FL, LG, GM, MH, HN, NE: Therefore each of the triangles EKF, FLG, GMH, HNE, is greater than half of the segment of the circle it stands in; because, if straight lines touching the circle be drawn through the points K, L, M, N, and the parallelograms upon the straight lines EF, FG, GH, HE he completed; each of the triangles EKF, FLG, GMH, HNE, shall be the half of the parallelogram in which it a 41. 1. is: But every segment is less than the parallelogram in which it is: Wherefore each of the triangles EKF, FLG, GMH, HNE, is greater than half the segment of the circle which contains it: And if these circumferences before named be divided each into two equal parts, and their extremities be joined by straight lines, by continuing to do

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this there will at length remain segments of the circle, which together shall be less than the excess of the circle EFGH, above the space S: Because, by the preceding Lemma, 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. Let then the segments EK, 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 EKFLGMHN is greater than the space S. Describe likewise in the circle ABCD the polygon AXBOCPDR similar to the polygon EKFLGMHN: As therefore the square of BD is to the square of FH, so is the polygon AXBOCPÒR to the 1. 12.

Book XII. polygon EKFLGMHN: But the square of BD is also to the square of FH, as the circle ABCD is to the space S: 11.5. Therefore as the circle ABCD is to the space S, so is the

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polygon AXBOCPDR to the polygon EKFLGMHN: But the circle ABCD is greater than the polygon contained 14. 5. in it; wherefore the space S is greater than the polygon EKFLGMHN: But it is likewise less, as has been demonstrated; which is impossible. Therefore the square of BD is not to the square of FH, as the circle ABCD is to any space less than the circle EFGH. In the same manner, it may be demonstrated, that neither is the square of FH to the square of BD, as the circle EFGH is to any space less than the circle ABCD. Nor is the square of BD to the square of FH, as the circle ABCD is to any space greater than the circle EFGH: For, if possible, let it be so to T, a space greater than the circle EFGH: Therefore, inversely, as the square of FH to the square of BD, so is the space T to the circle ABCD. But

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as the space+ T is to the circle ABCD, so is the circle EFGH to some space, which must be less than the circle ABCD, because the space T is greater, by hypothesis, than the circle EFGH. Therefore as the square of FH is to

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 propositious.

the square of BD, so is the circle EFGH to a space less Book XII. than the circle ABCD, which has been demonstrated to be impossible: Therefore the square of BD is not to the square of FH as the circle ABCD is to any space greater than the circle EFGH And it has been demonstrated, that neither is the square of BD to the square of FH, as the circle ABCD to any space less than the circle EFGH: Wherefore, as the square of BD to the square of FH, so is the circle ABCD to the circle EFGH†. Circles therefore are, &c. Q.E.D. a

PROP. III. THEOR.

EVERY pyramid having a triangular base, may be See N. divided into two equal and similar pyramids having triangular bases, and which are similar to the whole pyramid, and into two equal prisms which together are greater than half of the whole pyramid.

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Let there be a pyramid of which the base is the triangle ABC and its vertex the point D: The pyramid ABCD may be divided into two equal and similar pyramids having triangular bases, and similar to the whole; and into two equal prisms which together are greater than half of the whole pyramid.

Divide AB, BC, CA, AD, DB, DC, each into two equal parts in the points E, F, G, H, K, L, and join EH, EG, K GH, HK, KL, LH, EK, KF, FG. Because AE is equal to EB, and AH to HD, HE is parallela to DB: For the same reason, HK is parallel to AB: E Therefore HEBK is a parallelogram, and HK equal to EB: but EB is equal to AE; therefore also AE is equal

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to HK: And AH is equal to HD; wherefore EA,AH, are equal to KH, HD, each to each; and the angle EAH

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is equal to the angle KHD; therefore the base EH is 29. 1.

+ Because, as a fourth proportional to the squares of ED, FH, and the circle ABCD, is possible, and that it can neither be less nor greater than the circle EFGH, it must be equal to it.

J 4. 1.

Book XII. equal to the base KD, and the triangle AEH equal and similar to the triangle HKD: For the same reason, the triangle AGH is equal and similar to the triangle HLD: And because the two straight lines EH, HG, which meet one another, are parallel to KD, DL, that meet one an other, and are not in the same plane with them, they contain 10. 11. equal angles; therefore the angle EHG is equal to the angle KDL. Again, because EH, HG, are equal to KD, DL, each to each, and the angle EHG equal to the angle KDL; therefore the base EG is equal to the base KL: And the triangle EHG equald and similar to the triangle KDL: For the same reason, the triangle AEG is also equal and similar to the triangle HKL. Therefore the pyramid, of which the base is the triangle AEG, and of which the C. 11. vertex is the point H, is equal and similar to the pyramid, the base of which is the triangle KHL, and vertex the point D. And because

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HK is parallel to AB, a side of the tri-or-Nun angle ADB, the triangle ADB is equiangular to the triangle HDK, and their 4. 6. sides are proportionals. Therefore the triangle ADB is similar to the triangle HDK: And for the same reason, the triangle DBC is similar to the triangle DKL; and the triangle ADC to the triangle HDL; and also the triangle ABC to the triangle AEG: But the E triangle AEG is similar to the triangle HKL, as before was proved; therefore

21. 6. the triangle ABC is similar to the tri

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angle HKL. And the pyramid of which the base is the B. 11. & triangle ABC, and vertex the point D, is therefore similari 11 Def. 11. to the pyramid of which the base is the triangle HKL, and vertex the same point D: But the pyramid of which the base is the triangle HKL, and vertex the point D, is similar, as has been proved, to the pyramid the base of which is the triangle AEG, and vertex the point H: Wherefore the pyramid, the base of which is the triangle ABC, and vertex the point D, is similar to the pyramid of which the base is the triangle AEG, and vertex H: Therefore each of the pyramids AEGH, HKLD is similar to the whole. pyramid ABCD: And because BF is equal to FC, the pa* 41. 1. rallelogram FBEG is doublek of the triangle GFC: But when there are two prisms of the same altitude, of which one has a parallelogram for its base, and the other a triangle

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