ELEMENTS.

BOOK VI.

DEFINITION S.

1. SIMILAR Right-lined Figures are fuch as bave each of their several Angles equal to one anoiber, and the Sides about the equal Angles proportional to each other.

II. Figures are faid to be reciprocal, when the antecedent and confequent Terms of the Ratios are in each Figure.

III. A Right Line is faid to be cut into mean and extreme Proportion, when the Whole is to the greater Segment, as the greater Segment is to the leffer.

IV. The Altitude of any Figure is a perpendicular Line drawn from the Top, or Vertex, to the Bafe.

V. A Ratio is faid to be compounded of Ratios, when the Quantities of the Ratios, being multiplied into one another, do produce a Ratio.

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#38. 1.

+ Dif. 5.

PROPOSITION I.

THEOREM.

Triangles and Parallelograms that have the fame
Altitude, are to each other as their Bafes.

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ET the Triangles A B C, A C D, and the Parallelograms EC, C F, have the fame Altitude, viz. the Perpendicular drawn from the Point A to B.D. Lfay, as the Bafe B C is to the Base CD, fo is the Triangle ABC, to the Triangle A CD; and fo is the Parallelogram E C to the Parallelogram CF.

For, produce ED both Ways to the Points H and L; and take G B, GH, any Number of Times equal to the Bale BC; and D K, KL, any Number of Times equal to the Base CD; and join A H, A G, AK, AL.

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Then, because CB, BG, GH, are equal to one another, the Triangles A H G, A G.B, A B.C, also, will be equal to one another: Therefore the fame Multiple that the Bafe HC is of the fame BC, fhall the Triangle AHC be of the Triangle ABC. By the fame Reafon, the fame Multiple that the Bafe LC is of the Bafe CD, fhall the Triangle ALC be of the Triangle AD: And if the Bate HC be equal to the Bafe CL, the Triangle AHC is alfo * equal to the Triangle A LC: And if the Bafe H C exceeds the Bafe CL then the Triangle AHC will exceed the Triangle ALC. And if the Bafe HC be lefs than CL, then the Triangle AHC will be lefs than ALC. Therefore, fince there are four Magnitudes, viz. the two Bates BC, CD, and the two Triangles ABC, ACD; and fince the Bafe H.C, and the Triangle A H C, are Equimultiples of the Base B C, and the Triangle ABC: And the Bafe CL, and the Triangle A HC, are Equimultiples of the Bafe C D, and the Triangle-A DC: And it has been proved, that if the Bafe HC exceeds the Bafe CL, the Triangle AHC will exceed the Triangle ALC; and if equal, equal; if lefs, less: Therefore, as the Bale BC is to the Base CD, fo† is the Triangle ABC to the Triangle A CD.

And

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And because the Parallelogram EC is + double to † 41. 1. the Triangle ABC; and the Parallelogram FC, double + to the Triangle A C D and Parts have the fame Proportion* as their like Multiples: There- * 15. 1. fore, as the Triangle ABC is to the Triangle A CD, fo is the Parallelogram E C to the Parallelogram C F. And fo, fince it has been proved that the Bafe BC is to the Base CD, as the Triangle A B C is to the Triangle A CD; and the Triangle ABC is to the Triangle A CD, as the Parallelogram EC is to the Parallelogram CF; it fhall be ‡, as the Bafe B C is 1 11. 1. to the Base C D fo is the Parallelogram E C to the Parallelogram F C. Wherefore, Triangles and Parallelograms, that have the fame Altitude, are to each other as their Bafes; which was to be demonstrated.

PROPOSITION II.
THEOREM.

If a Right Line be drawn parallel to one of the
Sides of a Triangle, it shall cut the Sides of the
Triangle proportionally; and if the Sides of the
Triangle be cut proportionally, then a Right
Line, joining the Points of Section, shall be
parallel to the other Side of the Triangle.

LET DE be drawn parallel to B C, a Side of the
Triangle ABC. I fay, D B is to D A, as CE
is to E A.

For, let B E, CD be joined.

Then the Triangle BDE is equal to the Tri- * angle CDE; for they ftand upon the fame Base DE, and are between the fame Parallels DE and BC; and A D E is some other Triangle. But equal Magnitudes have +. the fame Proportion to one and the fame Magnitude. Therefore, as the Triangle BDE is to the Triangle A DE, fo is the Triangle CDE to the Triangle A D E.

37. I.

But as the Triangle BDE is to the Triangle ADE, fo is BD to DA; for fince they have the of this. fame Altitude, viz. a Perpendicular drawn from the Point E to A B, they are to each other as their Bafes. And, for the fame Reason, as the Triangle C D E is to

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the

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the Triangle ADE, fo is C E to E A: And there fore as B D is to D A, fo is CE to E A.

And if the Sides A B, A C, of the Triangle A B C, be cut proportionally; that is, fo that BD be to DA, as CE is to EA; and if DE be joined; I say, DE is parallel to BC.

For, the fame Conftruction,remaining, because B D is to DA, as CE is to EA; and B D is + to DA, as the Triangle BDE is to the Triangle ADE; and CE is to E A, as the Triangle C D E is to the Triangle A DE; it fhall be as the Triangle B D E is to the Triangle ADE, fo is the Triangle C DE to the Triangle AD E. And fince the Triangle B D E, CDE have the fame Proportion to the Triangle AD E, the Triangle B DE fhall be † equal to the Triangle CDE; and they have the fame Bafe DE: But equal Triangles being upon the fame Bafe, ‡ are between the fame Parallels; therefore D E is a parallel to B C. Wherefore, if a Right Line be drawn parallel to one of the Sides of a Triangle, it shall cut the Sides of the Triangle proportionally; and if the Sides of the Triangle be cut proportionally, then a Right Line joining the Points of Section shall be parallel to the other Side of the Triangle; which was to be demonftrated.

PROPOSITION III.

THEOREM.

If one Angle of a Triangle be bisected, and the Right Line that bifects the Angle, cuts the Bafe alfo; then the Segments of the Bafe will bave the fame Proportion as the other Sides of the Triangle. And if the Segments of the Bafe bave the fame Proportion that the other Sides of the Triangle have; then a Right Line, drawn from the Vertex, to the Point of Section of the Bafe, will bifet the Angle of the Triangle.

LET there be a Triangle A B C, and let its Angle BAC be bifected by the Right Line A D. I fay, as BD is to D C, fo is B A to AC,

For,

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For, through C draw CE parallel to DA, and produce B A, till it meets C E in the Point E.

37. z.

Then, because the Right Line AC falls on the Parallels AD, EC, the Angle ACE will be ‡ equal ‡ 29. 1. to the Angle CAD: But the Angle CAD (by the Hypothefis) is equal to the Angle BAD. Therefore the Angle BAD will be equal to the Angle ACE. Again, because the Right Line B A E falls on the Parallels A D, E C, the outward Angle B A D is + equal +.7.5. to the inward Angle A EC; but the Angle ACE has been proved equal to the Angle BAD: Therefore ACE fhall be equal to A EC; and fo the Side AE is equal to the Side A C. And because the Line A D is drawn parallel to CE, the Side of the Triangle BCE, it fhall be, as B Dis to DC, fo is * 2 of this. BA to A E; but A E is equal to A C. Therefore, as B D is to DC, fo is + BA to A C.

And if BD be to DC, as BA is to A C; and the Right Line AD be joined; then, I fay, the Angle BAC is bifected by the Right Line A D.

For, the fame Conftruction remaining, because BD

6.1.

is to D C, as BA is to A C; and as BD is to DC, fo is BA to AE; for AD is drawn parallel to one † 2 of this. ‡ Side E C of the Triangle B CE; it fhall be, as BA

26. 1.

is to A C, fo is BA to A E. Therefore AC is equal to AE+; and, accordingly, the Angle A E C is equal + 9.4. to the Angle ECA: But the Angle A E C is equal *to the outward Angle BAD; and the Angle ACE equal to the alternate Angle CAD. Wherefore the Angle BAD is alfo equal to the Angle CAD; and fo the Angle BAC is bifected by the Right Line A D. Therefore, if one Angle of a Triangle be bifected, and the Right Line, that bifects the Angle, cuts the Bafe alfo; then the Segments of the Bafe will have the fame Proportion as the other Sides of the Triangle. And if the Segments of the Bafe have the fame Proportion that the other Sides of the Triangle have; then a Right Line, drawn from the Vertex, to the Point of Section of the Bafe, will bifect the Angle of the Triangle ; which was to be demonstrated.

PRO