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E LE M E N T S.

BOOK

VI.

DEFINITION S. 1. SIMILAR Right-lined Figures are such 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 said to be reciprocal, when the

antecedent and consequent Terms of the Ratios

are in each Figure. III. A Right Line is said to be cut into mean and

extreme Proportion, when the whole is to the greater Segment, as tbe greater Segment is to

the lefser. IV. The Altitude of any Figure is a perpendicu

lar Line drawn from the Top, or Vertex, to

the Base. V. A Ratio is said to be compounded of Ratios,

wben the Quantities of the Ratios, being multiplied into one another, de produce a Ratio.

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PROPOSITION I.

THEOREM.
Triangles and Parallelograms that have the same

Alitude, are to each other as tbeir Bases.
ET the Triangles A B C, AC D, and the Pa-
rallelograms EC, CF, have the same Altitude,

viz. the Perpendicular drawn from the Point A
to. B.D. I say., as the Base B C is to the Bafe CD,
so is the Triangle A B C, to the Triangle A CD;
and so is the Parallelegram E C to the Parallelogram
CF.

Eor, produce E D both Ways to the Points H and
L; and take GB, GH, any Number of Times equal
10 the Bale B C ; and D K, KL, any Number of
Times equal to the Bale CD; and join A H, AG,
AK, AL.

Then, because CB, BG, GH, are equal to one

another, the Triangles AH G, A G.B, AB.C, also, 38. 1.

will be * equal to one another : Therefore the fame
Multiple that the Base HC is of the fame BC, thall
the Triangle AHC be of the Triangle ABC. By
the same Reason, the fame Multiple :hat the Base LC
is of the Base CD, fhall the Triangle ALC be of
the Triangle A CD. And if the Bale HC be equal
to the Basc CL, the Triangle A HC is also * equal
to the Triangle ALC: And if the Base H C exceeds
the Base CL then the Triangle A H C will exceed
the Triangle ALC. And if the Base HC be less
than CL, then the Triangle A HC will be less than
ALC. Therefore, since there are four Magnitudes,
viz, the two Bales BC, CD, and the two. Triangles
ABC, ACD; and, fince the Base H.C, and the
Triangle AHC, are Equimultiples of the Base B C,
and the Triangle ABC: And the Base CL, and the
Triangle A H°C, are Equimultiples of the Bale CD,
and the Triangle-A DC: And it has been proved,
that if the Bale H C exceeds the Base C L, the Tri-
angle AHC will exceed the Triangle ALC; and if

equal, equal; if less, less : Therefore, as the Base + Def. 5. B C is 'to the Base CD, fo + is the Triangle ABC to the Triangle ACD.

And

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And because the Parallelogram EC is + double to t 41. t. the Triangle A BC; and the Parallelogram FC, double † to the Triangle A CD: and Parts have the same Proportion * as their like Multiples : There. 15.1. fore, as the Triangle A B-C is to the Triangle ACD, fo is the Parallelogram EC to the Parallelogram CF. And-so, since it has been proved that the Base B C 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 Parallelgram EC is to the Parallelogram CF; it shall be I, as the Base B C is 1 11. 1. to the Bale CD so is the Parallelogram EC to the Parallelogram FC. Wherefore, Triangles and Parallelograms, that have the same Altitude, are to each other as their Bases; which was to be demonstrated.

PROPOSITION II.

THEOREM.
If a Right Line be drawn parallel to one of the

Sides of a Triangle, it fall 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, Mall be
parallel to the other Side of the Triangle.
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ET D E be drawn parallel to B C, a Side of the

Triangle A BC. I say, D B is to D A, as C E is to E A.

For, let B E, C D be joined.

Then the Triangle B D E is * equal to the Tri- * 37.1. angle CDE; for they stand upon the fame Base DE, and are between the same Parallels D E and BC; and A D E is some other Triangle. But equal Magnitudes have the same Proportion to one and the fame Magnitude. Therefore, as the Triangle BDE is to the Triangle A DE, so is the Triangle CDE 10 the Triangle A DE.

But as the Triangle B D E is to the Triangle ADE, so I is B D to DA; for fince they have the 11 of ibis. fame Altitude, viz. a Perpendicular drawn from the Point E to A B, they are to each other as their Bases. And, for the fame Reason, as the Triangle CDE is to

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

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

For, the same Construction remaining, because BD is to DA, as CE is to EA; and B D is † to DA, as the Triangle B D E is to the Triangle ADE; and C E is to E A, as the Triangle C D E is to the Triangle A DE; it shall ve as the Triangle B D E is to the Triangle A D E, so is * the Triangle C DE to the Triangle A D E. And since the Triangle B DE, CDE have the same Proportion to the Triangle AD E, the Triangle BDE shall be + equal to the Triangle CDE; and they have the same Base DE: But equal Triangles being upon the same Base, I are between the same 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 obe 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 demonstrated.

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$39. 1.

PROPOSITION III.

THE OR EM.
If one Angle of a Triangle be bisierd, and the

Rigbı Line that bisečts the Angle, cuis the
Base also ; tben the Segments of the Base will
bave tbe same Proportion as the other Sides of
the Triangle. And if the Segments of be Base
bave the same Proportion that the orber Sides
of the Triangle bave ; tben a Right Line,
drawn from the Vertex, to the point of Section
of the Base, will bileet the Angle of ibe Trie
angle.

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ET there be a Triangle A B C, and let its Angle

BAC be * bifected by the Right Line A D. I say, as B D is to D C, fo is B A to A C.

For,

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For, through C draw * CE parallel to D A, and 31. 1. produce B A, till it meets C Ein the Point E.

Then, because the Right Line AC falls on the Parallels A D, EC, the Angle ACE will be f equal 1 29. . to the Angle CAD: But the Angle CAD (by the Hypothesis) is equal to the Angle B AD. 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 t.7.50 to the inward Angle A EC; but the Angle ACE has been proved equal to the Angle B AD: Therefore ACE shall be equal to AEC; and so the Side A E is equal I to the Side A C. And because the

16.1.
Line A D is drawn parallel to CE, the Side of the
Triangle BCE, it shall be, * as B Dis to DC, so is . 2 of this.
B A to A E; but A E is equal to A C. . Therefore,
as B D is to DC, so is + B A to AC.

And if B D be to DC, as B A is to A C; and the
Right Line A D be joined; then, I say, the Angle
BAC is bisected by the Right Line AD.

For, the same Construction remaining, because BD
is to D C, as B A is to A C; and as B D is to DC,
so is I В A to A E; for A D is drawn parallel to one I 2 of ibis.
Side E C of the Triangle BCE; it shall be, as B A
is to A C, so is B A to A E. Therefore A C is equal
to AE t; and, accordingly, the Angle A E C is equal + 9.4.
to the Angle E CA: But the Angle A E C is equal

to the outward Angle BAD; and the Angle 26. 1. ACE equal to the alternate Angle CAD. Where, fore the Angle BAD is also equal to the Angle CAD; and so the Angle BAC is bisected 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 Base also ; then the Segments of the Base will have the same proportion as the other sides of the Triangle. And if the Segments of the Bafe have the same Proportion that the other Sides of the Triangle have ; then a Right Line, drawn from the Vertex, to the point of Section of the Base, will bisect the Angle of the Triangle ; which was to be demonstrated.

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