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Book VI.

PROP. I.

THE OR.

T
TRIANGLES and parallelogranis that have the same altitude,

are to each other as their bajes.

Let the triangles ABC, ACD, and the parallelograms EB, FD have the same 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 à 33. 1. another a ; and ACD, ADK, AKL, ALM equal to one

another a ; then, because HC is taken any multiple of BC, and CM any other multiple of CD, the triangle AHC is the same 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, less; therefore ABC b Def. 5.5. is to ACD as BC is to CD) b; 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.

C41. I.

d 15. 5. e il. 5.

PRO P. II. THEO R.

I
F a right line be drawn parallel to one of the sides of a triangle,

it will cut the other hdes proportionally; and if a line cut the two sides of a triangle proportionally, that right line fall be parallel to the other side of the triangle.

2 37. I.

b 7. 5. CI. dil. 5.

Let DE be drawn parallel to BC, one side 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 fome other triangle; therefore BDE is to ADE as DEC is to ADE 0 : But BDE is to ADE as BD is to ADC; and DEC is to ADE as EC is to AE~; therefore BD is to DA as CE is to EA d; and if BD is to DA as CE is to EA, then DE is parallel to BC.

For the same construction remains : As BD is to DA, so 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 d ; therefore the triangles BDE, CDE are equal; therefoie DE is parallel to BC f, Wherefore, &c.

e 9. 5. f 39. 1.

Book VI.

PRO P. III. THEOR.

Fone angle of a triangle be biseEted by a right line, which like

have the same proportion to one another that the other sides of the triangle have: And if the segments of that fide have the same proportion to one another that the other sides of the triangle have, then a right line, drawn from the point of section to the vertex, will bifect the opposite angle.

Let there be a triangle ABC, and let one of its angles, as BAC, be bisected by the right line AD, then, as BD is to DC, so 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 ACEB: But the angle BAD is equal to CAD"; b 29. 1.

с Нур. therefore the angle BAD is equal to ACE 4: But BAD is equal to AEC ; therefore AEC is equal to ACEd; therefore AE is d Ax. I. 1. equal to AC4: 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.

For the same construction remains; BD is to DC as BA is to AC°; and BD is to DC as 6A is to AEf; therefore BA is to AC as BA is to AE%; therefore AE is equal to ACh; there- & 11.5. fore the angles ACE, AEC are equali: But ACE is equal to DACO; therefore the angle AEC is equal to DAC 4: But AEC is equal to BADb; therefore BAD is equal to DAC4; therefore the angle BAC is bifected by the right line AD. Wherefore, &c.

h 9.5.

i

5. I.

PRO P. IV. THE O R.

HE fides about the equal angles of equiangular triangles

are proportional; and the sides subtending 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 sides that are about the equal angles are proportional, and the sides subtending the equal angles homologous, or of like ratio.

Let

1

a 17. I.

Book VI. Let the sides BC, CE be placed in the same right line; then,

because the two angles ABC, ACB are less than two right angles", and the angles ACB, DEC are equal, the angles

ABC, DEC are less than two right ones : The right lines AB, b Cor. 17. DE being produced, will meet o in some point; which let be F;

then, because the angle DCE is equal to ABC, DC is parallel to C 28. 1. ABC: But the angle BAC is equal to CDE d; and BAC to d dyp. 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 CE5: Alter. BA is to BC as AF or CD is to CE. Again, because h 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 DEI. Wherefore, &c.

€ 29. I. f 27. I.

8 34 I, h 2.

i 22. S.

PROP. V. and VI. THE O R.

If the

sides of two triangles are proportional, or if one angle of the equal angles proportional, the triangles will be equiangulur, and the angles which the homologous fides fubtend will be 6qual.

Let there be two triangles ABC, DEF, having their fides proportional; viz. AB to BC as DE to EF; and as BC is to CÀ, so is EF to FD; and as BA to AC, so 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 subtend are equal.

First, Let the triangles ABC, DEF have their sides proportional (fig. 1.); at the points E, F, with the right line EF; make the angle FEG equal to the angle ABC ~; and EFG to ACB; then

the remaining angles EGF, BAC will be equal b; and because Cor. 32. the two triangles ABC, EFG are equiangular, AB is to BC as

GE to EFC: But as AB is to BC lo is DE to EFd; therefore d Hyp. DE is to EF as GE is to EF €; therefore DE is equal to Gif;

for the same reason FG is equal to FDf, and EF is common; f 9. S.

therefore the triangles DEF, GEF are equal 6; 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 likewise equiangular to ABC.

a 23. 1. h

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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 w right line DF, make the angle FDG equal to the angle EDF, or BẠC; and the angle DFG equal to the angle ACB ; then the remaining angles at G, B, are equal "; because BA is to AC as h cor, 32. I. ED to DF ; and likewise, as GD is to DF; therefore DG is equal to DEf. For the same reason EF is equal to FG; and f 9. S. the triangles DEF, DFG, equiangular ; but DFG is equiangular to ABC; therefore DEF is likewise equiangular to ABC. Wherefore, &c,

PRO P. VII. THE O R.

IF there are two triangles, having one angle of the one equal

to one angle of the other, and the sides about a second angle of the one proportional to the sides about the correspondent angle of the other, and the remainimg third angles, either both less, or both not less than right angles ; then the triangles will be equiang "lar, and have these angles equal, about which the sides are proportional,

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

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 sides 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 BCR, and the angle BCG to BGC f; there- 9.5. fore BGC, BCG, are each less than a right angle; therefore DFE is less than a right angle 8; but BGA, BGC, are e- & Hyp. qual to two right angles ", and BGC is proved less than a right h 13. angle; therefore BGA is greater than a right angle ; therefore DFE is likewise greater than a right angles, and less; which is impoffible; therefore ABG is not equal to DEF; nor can any angle but ABC be equal to DEF. Wherefore, &c.

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PROP.

Book VI.

PRO P. VIII. THEOR.

If a perpendicular be drawn in a right angled triangle

, from the right angle to the base, then the triangles on each side of the perpendicular will be fimilar to the whole, and to one another.

1.

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 similar 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 14.

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 DCB; they are likewise fimilar to one another; for 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 fimi¢ Defo 1. lar to the wholeç, and to one another. Wherefore, &c.

Cor. Hence, in a right angled triangle, if a perpendicular is let fall from the right angle to the base, that perpendicular is a mean proportional to the segments of the base; and each of the sides containing the right angle is a mean proportional to the whole bafe, and that fegment next to the side.

PRO P. IX. PROB.

T

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

2 3. J. b 31. 1.

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; assume any point D in the line AC; and make DE, EC, each equal AD , join BC; and through D draw DF parallel to BC 6.

Then, because FD is parallel to BC, a side 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.

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