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Wherefore a AFCG has been described, which Om ABCD, and having the L FCG = gn. L E.-Q. E. F.

PROP. XLIII. THEOR.

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GEN. ENUN.—The complements of the parallelograms, which are about the diameter of any parallelogram, are equal to one another.

PART. Enun.—Let ABCD be a om, of which the diam. is AC; EH, FG the oms about ac, i.e. through which ac passes ; and BK, KD the other o ms which make ир the whole figure ABCD, and B ... called complements. Then the complement BK = complement KD.

DEMONST.-1. Because the diam. Ac bisects the Om ABCD (Prop. XXXIV.), .-. the ABC

A ADC. 2. For the same reason, the A AEK = A AHK, and the A KGC = A KFC.

3. .. the complement Bk = A ABC – ( AEK + AKGC) A ADC (AHK + s KFC) = complement kb.

Wherefore the complements &c.—Q. E. D. There are other cases of this

Fig. 1. Proposition ; since it is not necessary that the ams about AC should be united at k, or that the complements should be ms. The Theorem may be equally proved, when the ams EH, FG are not connected

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To a given straight line to apply a parallelogram, which shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle.

PART. ENUN.- Let AB be the gn. st. line,

gn. rectilineal Z; then it is required to apply to the st. line AB a o A c, and having an L = D.

Const. and DEMONST.—Make the om BEFG = A c, and having the Z EBG = L D (Prop. XLII.), and let BE be in the same st. line with AB; through a draw au || to bg or EF (Prop. XXXI.), and meeting fg produced in H; join hb.

2. And because the st. line hf falls upon the || st. line AH, EF, . . the < $ AHF + HFE = 2 rt. Z $ (Prop. XXIX.), .. the 2 bhf + HFE are < 2 rt. Zs; .. HB, FE will meet if produced towards B and E. (Prop. XXIX. Cor.)

3. Let them meet in K; through a draw KL Il to EA or fh (Prop. XXXI.), and meeting HA produced in L; also produce go to m.

4. Then HLKF is a om, of which the diam. is HK; AG, ME are the oms about hk, and LB, BF are the complements : .. the complement LB = complément BF (Prop. III.) = A C. (Const.)

5. Also the ABM = vertical Z EBG (Prop. XV.) = 2 D. (Const.)

Wherefore the om LB has been applied to the

gn. st. line AB, whieh the having the Z ABM = gn. < D.—Q.E.D.

The converse Problem may be thus effected.

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PROP. AAA. PROB. GEN. ENUN.-To a given straight line to apply a tri. angle, which shall be equal to a given parallelogram, and having an angle equal to a given rectilineal angle.

PART. ENUN.-Let AB be the gn. st. line, CDEF the gn. Om, G the gn. rectilineal L. Then it is required to apply to AB a A = Om CDEF, and having an L= L G.

Const.–Draw the diam. CF; produce EF to H, making FH = join ch. To the st. line AB apply the' om ABKL ДcЕН, and having the L ABK = LG (Prop. XLIV.); produce bk to m, making KM = BK, and join AM, AK; then A ABM is the A required.

DMONST.–For the A ABM = 2 A ABK (Prop. NN.) = 0 BL (Prop. XLI.) = A CEH (Const.) = 2 Δ CER (Prop. NN.) OM CEFD. (Prop. XLI.) Wherefore to the gn. st. line AB has been applied a A

gn. Om ED, and has the ABM = G.-Q. E. F.

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PROP. XLV. PROB.

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GEN. ENUN.To describe a parallelogram equal to a given rectilineal figure, and having an angle equal to a given rectilineal angle.

PART. ENUN.—Let ABCD be the gn. rectilineal figure, E the gn. rectilineal Z; then it is required to describe a om= ABCD, and having an

= LE. CONST. — Join DB; describe the o MFH = A ADB, and having the Z FKH = 2 E (Prop. XLII.); to the st. line gh apply the om GM =

DBC, and having < ghm = L E (Prop. XLIV.); then the figure FKML is the om required.

DEMONST.-1. Since the 2 FKH = LE= GHM (Const.), add to each side the < KHG, .:. the ZS FKH + KHG ZS KHG + GHM (Ax. 2).

2. But the ZS FKH + KHG = 2 rt. Zs (Prop. XXIX.); .. the $ KHG + GHM = 2 rt. Zs (Ax. 1).

3. And because at the pt. h in the st. line gh, the two st. lines KH, Hm upon the

opposite side of it, makes the adjacent z $ = 2 rt. Zs; .. ku is in the same st. line with HM. (Prop. XIV.)

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4. Again, because the st. line gh meets the Il st. lines KM, FG,

the alternate z MHG = alternate 2 HGF. (Prop. XXIX.)

5. To each add the 2 ugl, .'. the SMHG + url = _$ HGF + HGL (Ax. 2).

6. But the ZS MHG HGL = 2 rt. Zs (Prop. XXIX.), .. the <$ HGF + HGL = 2 rt. 2 (Ax. 1), and ... FG is in the same st. line with gl. (Prop. XIV.)

7. Now, because ky is || to hg, and ug to ML, .'. KF is || to ML (Prop. XXX.); and KM is also || to FL; .. KFLM is a om. 8. Hence the figure ABCD A ADB +

om FH + Om HL (Const.) m KFLM.

Wherefore a om has been described = to the gn. figure ABCD, and having the 2 FKM = gn. < E.-Q. E. F.

COR,-From this it is manifest how to a given st. line to apply a om, which shall have

= a gn. rectilineal Z, and be = a gn.' rectilineal figure: viz. by applying to the st. line a om = to the first ABD, and having an 2 = the gn. L.

It is also manifest how to proceed, whatever be the number of sides to the gn. rectilineal figure. For it may be divided into as many As, except two, as the figure has sides ; and a am= to each of them in succession is to be applied to the opposite side of each successive am.

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