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Bisect AB in E (a); upon EB describe the parallelogram EL, similar to D (b); make the parallelogram GH equal to EL and CC both (c), and similar to D, or EL (d). KH be the side homologous to FL, and KG to FE: and because GH exceeds EL, its sides KH, KG exceed the sides FL, FE. Make FM equal to KH, and FN equal to KG; and complete the scheme MNA.

Let

MN is equal and similar to GH; and therefore similar to EL; and so, MN, EL are about the same diameter FX (e). Therefore MN being equal to GH, is equal to EL and CC both: and its gnomonic part NOL, is therefore equal to the triangle CC. But NA equals NB (f), or its equal BM (g); therefore the gnomon NOL, or its equal CC, is equal to the parallelogram AX; which is applied to AB; having its excess OP similar to EL, since they are about the same diameter (h); and EL is similar to D; wherefore OP is also similar to D, as required.

Therefore to a given straight line is applied, &c., which was to be done.

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The four following Problems are taken from Simson's Notes: Note 1. To apply a defective rectangle equal to a given square to a given straight line, the defect to be a square but the given square must not exceed that upon half the given line.

Given the straight line AB; also the square of C, equal to the rectangle to be applied, but not greater than the square of half AB.

Bisect AB in D; then if C equal AD, their squares are equal, and the defect is the square of DB.

But if C be less, for it may not be greater, than AD, draw DE at right

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angles to AB (a); make DE equal to C, and EF equal to AD, or DB; from E as centre, with the distance EF, describe an arc, meeting AB in G; upon GB describe the square BH (b); complete the rectangle AH, and join EG.

Now the rectangle AGXGB, with the square of DG, is equal to the square of AD, or its equal EG (c): but the square of EG equals the squares of ED, DG (d); therefore, the rectangle AGXGB, with the square of DG, equals the squares of ED and DG. Take from

both the square of DG: the rectangle AGXGB remains equal to the square of ED, or C.

Wherefore, to a given straight line AB, a rectangle is applied equal to a given square and defective by a square; which was to be done. (c) p. 5, 2;

Recite (a) p. 11, 1; (d) p. 47, 1.

(b) p. 46, 1;

Note 2. To apply an excessive rectangle, equal to a given square, to a given straight line; the excess to be a

square.

Given AB, a straight line, and the square of C, equal to the rectangle to be applied.

Bisect AB in D; draw BE at right angles to it, so that BE equal C; join DE, and with this as radius, describe a circle to meet AB produced in G; upon BG describe the square BH, and complete the rectangle BL.

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Since AB is bisected in D, and produced to G, the rectangle AGXGB, with the square of DB, is equal to the square of DG (a), or DE its equal, or the squares of DB and BE (b). Take from both sides the square of DB; therefore the rectangle AGXGB equals the square of BE, or its equal C. But the rectangle AGXGB is the rectangle AH; which is applied to AB, exceeding by the square BH, as required to be done.

Recite (a) p. 6, 2;

(b) p. 47, 1.

Note 3. To apply a defective rectangle to a given straight line, not greater than the square of its half, but equal to a given rectangle: the defect to be a square.

Given AB, a straight line; and CXD, a rectangle, equal to the defective one to be applied to AB, but not to exceed the square of half AB.

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On one side of AB, draw AE, BF at right K angles to it; so that AE-C, and BF=D: join EF, and bisect it in G: with GE radius, describe the circle EHF: join HF; make GK parallel to it, and GL parallel to AE.

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Since the angle EHF, in a semicircle, is equal to the right angle EAB, HF is parallel to AB, and AH to BF: wherefore AH-BF, and EAXAH EAXBF; that is, CXD.

And since EG-GF, and AE, LG, BF are parallels; therefore AL= LB, and EK KH (a): and since CXD may not exceed AL', or GK2, to each add KE2, therefore CXD+KE2; that is, EAXAH+ KH2, or AK2 (b), is not greater than GK2+KE2, that is, EG2 (c): consequently, AK, or GL, is not greater than EG.

Now, if GE be equal to GL, the circle will touch AB in L (d); and AL2 will equal EAXAH, or CXD, which is the thing required. But if GE, GL be unequal, GE is greater; and the circle will cut

AB in points M, N. Upon NB describe the square NO, and complete the rectangle NQ. Now since AL-BL, and ML=NL (a); therefore AM=NB; and ANXNB=ANXAM; that is, AEXAH (e), or CXD. But ANXNB is the rectangle AP; which is applied to AB, defective by the square NO, as was required to be done.

Recite (a) p. 3, 3;

(d) p. 36, 3;

(b) p. 6, 2;
(e) cor. p. 36, 3.

(c) p. 47, 1;

Note 4. To apply an excessive rectangle to a given straight line, equal to a given rectangle: the excess to be a square.

Given AB, a straight line; and CXD a rectangle equal to the excessive one to be applied to AB.

On the contrary sides of AB, at right angles to it, draw AE=C, and BF=D: join EF, and bisect it in G: with GE as radius, describe a circle EHF; join HF; draw GL parallel to AE; produce AB to meet the circle in points M, N; upon BN describe the square NO, and complete the rectangle BQ.

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N

The angle EHF, in a semicircle (a), equals the right angle EAB; therefore HF is parallel to AB (6), and AH=BF (c); therefore EAX AH EAXBF, or CXD.

And since ML=LN (d), and AL=LB; therefore MA-BN, and ANXNB AMX AN, or EAXAH (e); that is, CXD. But ANXNB is the rectangle AP, which is equal to the given rectangle CXD, and applied to AB, exceeding by the square NÕ, as required.

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Given AB, a straight line, to be cut into two such parts, AE, EB, as that AB is to AE as AE is to EB. Upon AB describe the square BC (a); and to AC, a A side of the square, apply CD, an excessive rectangle, equal to BC: the excess being the square AD. Note 4, above.

See

Then since BC=CD, and the part AF is common to both; removing this, the remainders AD and BF are C F equal: these figures are also equiangular; hence their sides are reciprocally proportional (b). Wherefore EF: ED:: AE: EB. But EF AC, or AB (c); and ÉD=AE. Therefore AB: AE:: AE: EB,

as stated. But AB is greater than AE, therefore AE is greater than EB (d). Hence the straight line AB is divided in extreme and mean ratio, as required (e).

Recite (a) p. 46, 1: (d) p. 14, 5;

(b) p. 14, 6; (e) def. 3, 6.

(c) p. 34, 1;

Otherwise. Divide AB in the point E; AEXAE may equal ABXEB (p. 11, 2): AB: AE AE: EB, &c., as above (p. 17, 6).

so that AE2, that is, then it will be that

31 Th. In right angled triangles, the rectilineal figure described upon the side subtending the right angle, is equal to the figures similar to it, described upon the sides containing the same angle.

Given the triangle ABC, right-angled at A; the rectilineal figure described upon BC, is equal to the figures similar to it, described upon AB, AC.

Draw AD perpendicular to BC: then the tri- B D angles ADB, CDA, are similar to CAB and to

each other (a).

1. Since CAB, ADB are similar. CB: AB::

AB: DB (b): the straight lines CB, AB, DB are therefore three proportionals; and CB: DB:: fig. upon CB: similar fig. upon AB (c). 2. Since CAB, CDA are similar, CB: CA :: CA: CD (b): the straight lines CB, CA, CD are therefore three proportionals; and CB: CD: fig. upon CB: similar fig. upon CA (c).

Taking these two proportions inversely (d):

DB: CB:: fig. upon AB: fig. upon CB; and

CD: CB fig. upon CA: fig. upon CB.

Adding 1st and 5th, 2d and 6th (e):

CD+DB: CB:: figs. upon CA+AB: fig. upon CB.

But CD+DB=CB; therefore figs. upon CA+AB-fig. upon CB (ƒ). Therefore, in right-angled triangles, &c.

Recite (a) p. 8, 6; (b) p. 4, 6:

Q. E. D.

(d) p. B, 5;

(e) p. 24, 5;

(c) cor. p. 19, 6;
(ƒ) p. A, 5.

32 Th. If two triangles, having two sides of the one proportional to two sides of the other, be joined at one angle, so as to have their homologous sides parallel to each other; their third sides shall be in one straight line.

Given two triangles ABC, DCE, joined at their A point C, with BA parallel to CD, and AC to DE; also BA: AC:: CD: DE-then BC, CE are in one straight line.

Because AC cuts the parallels BA, CD; and DC cuts the parallels AC, DE; the three alternate angles A, ACD, D are equal to each other (a).

And because the angles A, D are equal, and the B

sides about them proportionals, the triangles ABC, DCE are equiangular (b); therefore B=DCE, and ACB-E: moreover ACD=A. Therefore the angles ACB, ACD, DCE together, are equal to all the angles of the triangle ABC; that is, to two right angles (c); or half the compass of the angular point C: therefore BC, CE, form no angle with each other; but are in one straight line (d).

Wherefore, if two triangles, &c.
Recite (a) p. 29, 1; (b) p. 6, 6;

(d) cor. def. 11, 1; also p. 14, 1.

(c) p. 32, 1;

Q. E. D.

33 Th. In equal circles, angles, either at the centres or circumferences, have the same ratio as the arcs have on which they stand: so also have the sectors.

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equal to EF: join G to K, L, and H to M, N.

Then since the arcs BC, CK, KL are equal, their angles are equal (a); and the arc BL, and angle BGL, are equimultiples of the arc BC, and angle BGC.

And since the arcs EF, FM, MN are equal, their angles are equal (a); and the arc EN, and angle EHN, are equimultiples of the arc EF, and angle EHF.

If then the arc BL be equal to, greater, or less than the arc EN, the angle BGL is equal to, greater, or less than the angle EHN. BL and BGL being therefore equimultiples of BC the first, and BGC the third; and EN and EHN being equimultiples of EF the second, and EHF the fourth it follows (b), that BC: EF:: BGC: EHF. Therefore the central angles have the ratio of their arcs, as stated.

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But, because the central angles are second multiples of those at the circumference (c), BGC: EHF:: BAC: EDF (d); and so, BC: EF: BAC: EDF (e); that is, the angles at the circumference have also the ratio of their arcs.

The sectors also have the ratio of their arcs. Join C to B, K. Then since the radii GB, GC, GK are equal, and contain equal angles, as above, the bases are equal; namely, the chords BC, CK; finally, the triangles GBC, GCK are equal (ƒ). Moreover, the arcs were made equal, and the chords prove equal; therefore, the segments are similar and equal (g), the greater to the greater and the less to the less (h).

Now a sector, in this case, is composed of a triangle and one of the less segments; and these prove equal, each to each: therefore the sectors BGC, CGK are equal; and KGL may be proved equal to either, in the same way. Moreover, the sectors EHF, FHM, MHN may be proved equal to each other. Therefore, the sectors BGL,

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