« ForrigeFortsett »
E LE M E N T S.
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.
Alitude, are to each other as tbeir Bases.
viz. the Perpendicular drawn from the Point A
Eor, produce E D both Ways to the Points H and
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
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 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.
Sides of a Triangle, it fall cut the sides of the
Line, joining the Points of Section, Mall be
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
+ 1 of this.
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.
THE OR EM.
Rigbı Line that bisečts the Angle, cuis the
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, 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
And if B D be to DC, as B A is to A C; and the
For, the same Construction remaining, because BD
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.