Εικόνες σελίδας
PDF
Ηλεκτρ. έκδοση

Book VI.

2 33. 1.

PROP. I. THEOR.

RIANGLES and parallelograms that have the fame altitude, are to each other as their bafes.

TRIA

Let the triangles ABC, ACD, and the parallelograms EB, FD have the fame altitude, they are to one another as their bafes; viz. as BC to CD; for, produce BC both ways to H, M; take BG, GH each equal to BC; and DK, KL, LM each equal to CD; and join AG, AH, AK, AL, AM; then the triangles ABC, ABG, AGH are equal to one another; and ACD, ADK, AKL, ALM equal to one another; then, because HC is taken any multiple of BC, and CM any other multiple of CD, the triangle AHC is the fame multiple of the triangle ABC that HC is of BC; and the triangle CAM the fame multiple of CAD that CM is of CD: If HC be equal to CM, the triangle AHC will be equal to the triangle ACM; if greater, greater; and if lefs, lefs; therefore ABC b. Def. 5. 5. is to ACD as BC is to CD; but the parallelogram EB is double the triangle ABC, and FD double ACD; therefore the parallelogram EB is to the parallelogram DF, as the triangle ABC is to the triangle ACD; therefore the parallelograms EB, DF are to each other as their bases BC, CD. Wherefore, &c.

C 41. I.

d 15. 5.

e 11. 5.

2 37. I.

b 7. 5.

CI.

d 11. 5.

e g. 5.

f 39. I.

I'

PRO P. II. THE O R.

F a right line be drawn parallel to one of the fides of a triangle, it will cut the other hides proportionally; and if a line cut the two fides of a triangle proportionally, that right line shall be pa rallel to the other fide of the triangle.

Let DE be drawn parallel to BC, one fide of the triangle ABC, then AD will be to DB as AE is to EC; for join DC, BE, then the triangles BDE, DEC are equal; and ADE is some other triangle; therefore BDE is to ADE as DEC is to ADE: But BDE is to ADE as BD is to AD; and DEC is to ADE as EC is to AE; therefore BD is to DA as CE is to EAd; and if BD is to DA as CE is to EA, then DE is parallel to BC.

For the fame conftruction remains: As BD is to DA, fo is BDE to ADE; and CE is to EA as DEC is to ADE; therefore the triangle BDE is to the triangle ADE as DEC is to ADE; therefore the triangles BDE, CDE are equal; therefore DE is parallel to BC f. Wherefore, &c.

Book VI

PRO P. III. THE O R.

F one angle of a triangle be bifected by a right line, which likewife cuts the oppofite fide, then the fegments of that fide will have the fame proportion to one another that the other fides of the triangle have: And if the fegments of that fide have the fame proportion to one another that the other fides of the triangle have, then a right line, drawn from the point of fection to the vertex, will bifect the oppofite angle.

Let there be a triangle ABC, and let one of its angles, as BAC, be bifected by the right line AD; then, as BD is to DC, fo is BA to AC; for through C draw CE parallel to AD, a 31. 1. and produce BA till it meet CE in the point E; then, because AC falls on the parallels DA, CE, the angle DAC is equal to the angle ACE: But the angle BAD is equal to CAD; b 29. 1. therefore the angle BAD is equal to ACE: But BAD is equal to AEC b; therefore AEC is equal to ACE; therefore AE is d Ax. 1. 1. equal to AC: But BD is to DC as BA is to AEf; that is, to e 6. 1. AC; and if BD is to DC as BA is to AC, then the right line DA bifects the angle BAC.

c Hyp.

f 2.

h 9. 5. i 5. 1.

For the fame conftruction remains; BD is to DC as BA is to AC; and BD is to DC as BA is to AEf; therefore BA is to AC as BA is to AE; therefore AE is equal to AC1; there- g 11. 5. fore the angles ACE, AEC are equali: But ACE is equal to DAC; therefore the angle AEC is equal to DAC: But AEC is equal to BAD"; therefore BAD is equal to DACd; therefore the angle BAC is bifected by the right line AD. Wherefore, &c.

PRO P. IV. THE OR.

HE fides about the equal angles of equiangular triangles are proportional; and the fides fubtending the equal angles are homologous, or of like ratio.

Let the two equiangular triangles be ABC, DCE, viz. the angle ACB equal to the angle DEC; BAC to CDE; and ABC to DCE; then the fides that are about the equal angles are proportional, and the fides fubtending the equal angles homologous, or of like ratio.

Let

BOOK VI.

a 17. I.

I.

C 28. I. d Hyp.

Let the fides BC, CE be placed in the fame right line; then, because the two angles ABC, ACB are lefs than two right angles, and the angles ACB, DEC are equal, the angles ABC, DEC are lefs than two right ones: The right lines AB, b Cor. 17. DE being produced, will meet in fome point; which let be F; then, because the angle DCE is equal to ABC, DC is parallel to AB: But the angle BAC is equal to CDEd; and BAC to ACD; therefore the angles CDE, ACD are equal; therefore ED is parallel to ACf; therefore ACDF is a parallelogram, and FD is equal to AC, and AF to DC; and, because AC is parallel to FE, BA is to AF as BC is to CE: Alter. BA is to BC as AF or CD is to CE. Again, because BC is to CE as FD or AC is to DE, alter. BC is to AC as CE to ED: But AB is to BC as CD to CE; and BC to AC as CE to ED; therefore AB is to AC as CD to DE. Wherefore, &c.

e 29. I. f 27. I.

8 34 I.

n 2.

i 22. 5.

a 23. T.

[ocr errors]

PROP. V. and VI. THE OR.

IF the fides of two triangles are proportional, or if one angle of the one be equal to one angle of the other, and the fides about the equal angles proportional, the triangles will be equiangular, and the angles which the homologous fides fubtend will be equal.

Let there be two triangles ABC, DEF, having their fides proportional; viz. AB to BC as DE to EF; and as BC is to CA, fo is EF to FD; and as BA to AC, fo is ED to DF; or, if the angle BAC is equal to the angle EDF, and BA to AC as EĎ to DF; then the triangles ABC, EDF are equiangular, and the angle ABC equal to DEF; BCA to EFD, and BAC to 'EDF; that is, the angles which the homologous fides fubtend are equal.

Firft, Let the triangles ABC, DEF have their fides proportional (fig. 1.); at the points E, F, with the right line EF; make the angle FEG equal to the angle ABC; and LFG to ACB; then the remaining angles EGF, BAC will be equal b; and because h Cor. 32. the two triangles ABC, EFG are equiangular, AB is to BC as GE to EF: But as AB is to BC fo is DE to EFd; therefore DE is to EF as GE is to EF; therefore DE is equal to Gif; for the fame reafon FG is equal to FD f, and EF is common; therefore the triangles DEF, GEF are equal; and the remaining angles of the one equal to the remaining angles of the other, each to each: But the triangle EGF is equiangular to the triangle ABC; therefore DEF is likewife equiangular to ABC.

d Hyp.

e II. S. f 9. 5. g 8. I.

Secondly, Let the angle BAC be equal to the angle EDF, Book VI. (fig. 2.) and BA to AC as ED to DF; at the point D, with the right line DF, make the angle FDG equal to the angle EDF, or BAC; and the angle DFG equal to the angle ACB; then the remaining angles at G, B, are equal "; becaufe BA is to AC as h cor, 32. 1. ED to DF; and likewife, as GD is to DF; therefore DG is equal to DE f. For the fame reafon EF is equal to FG; and f 9. 5. the triangles DEF, DFG, equiangular; but DFG is equiangular to ABC; therefore DEF is likewise equiangular to ABC. Wherefore, &c,

IF

PRO P. VII. THE OR.

there are two triangles, having one angle of the one equal to one angle of the other, and the fides about a fecond angle of the one proportional to the fides about the correfpondent angle of the other, and the remaining third angles, either both less, or both not less than right angles; then the triangles will be equiangular, and have thefe angles equal, about which the fides are propor

tional.

Let the two triangles ABC, DEF, have an angle BAC in in the one equal to the angle EDF in the other; and the fides about the angles ABC, DEF, proportional, viz. AB to BC, as DE to EF; and the other angles at C, F, either both leis, or both not less than right angles; then the triangles ABC, DEF, are equiangular; the angle ABC equal to DEF, and ACB to DFE.

9.5.

f s T.

For, if the angle ABC be not equal to the angle DEF, let one of them, as ABC, be the greater, and make the angle ABG equal to the angle DEF; then the remaining angles a 23. 1. AGB, DFE', are equal, and the fides about the equal angles b cor. 32. 1. proportional, viz. AB to BG as DE to EF; but AB is to BC c 4. as DE to EF; therefore AB is to BG as AB is to BCd; there- d 11. s. fore BG is equal to BC, and the angle BCG to BGC f; there fore BGC, BCG, are each lefs than a right angle; therefore DFE is less than a right angles; but BGA, BGC, are e- 8 Hyp. qual to two right anglesh, and BGC is proved lefs than a right h 13. angle; therefore BGA is greater than a right angle; therefore DFE is likewife greater than a right angles, and lefs; which is impoffible; therefore ABG is not equal to DEF; nor can any angle but ABC be equal to DEF. Wherefore, &c.

PROP

Book VI.

J.

b4.

PRO P. VIII. THE OR.

IF a perpendicular be drawn in a right angled triangle, from the right angle to the base, then the triangles on each fide of the perpendicular will be fimilar to the whole, and to one another.

Let ABC be a right angled triangle; and from the point A of the right angle BAC, let fall the right line AD perpendicular to the base BC; then the triangles ABD, ADC, are fimilar to ABC, and to one another.

For the right angles BAC, ADB, are equal; and the angle at B common to the two triangles ABC, ABD; therefore the a cor. 32. remaining third angles C and BAD are equal; therefore BC is to BA as BA is to BD. Again, because the right angles BAC, ADC, are equal, and C common to both, the remaining third angles B, DAC, are equal; therefore BC is to AC as AC is to DC; they are likewife fimilar to one another; the angles ADC, ADB, are each right angles, and the angle C equal to BAD, and B to DAC; therefore BD is to AD as AD is to DC; therefore the triangles ADB, ADC, are fimilar to the whole, and to one another. Wherefore, &c.

c Def. 1.

for

COR. Hence, in a right angled triangle, if a perpendicular is let fall from the right angle to the bafe, that perpendicular is a mean proportional to the fegments of the bafe; and each of the fides containing the right angle is a mean proportional to the whole bafe, and that fegment next to the fide.

PROP. IX. PROB.

[merged small][merged small][ocr errors]

T

O cut off any part required from a given right line.

Let AB be a given right line, it is required to cut off any part of it, as one third.

From the point A draw any right line AC, making any angle with the line AB; affume any point D in the line AC; and make DE, EC, each equal AD 2; join BC; and through D draw DF parallel to BC.

Then, becaufe FD is parallel to BC, a fide of the triangle ABC, AF is to FB as AD is to DC; but AD is one third part of AC; therefore AF is one third part of AB; which was required. Wherefore, &c.

« ΠροηγούμενηΣυνέχεια »