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alfo at C and E, be Equal one to another, each to its Correfpondent I fay, the Sides about the Equal Angles are Proporti onal: For,
1. As A B 2. As A B 3. As AC
is to BC:: So is AD: to DE.
is to CB: So is AE to ED.
(2.) Because the Angles B A C and D A E are equal, therefore if A B be applied to A D, then A C fhall of Neceffity fall in AE; and by fuch Application fhall fuch a Figure be made.
In which Figure, because that A B and AC do meet together, and alfo the Angles at B and D, and at C and E, are Equal; therefore the other Sides B C and DE fhall be Parallel, (by the firft bereof). But in a plain Triangle a Right Line, Parallel to the Bafe, cutteth the Sides proportionally (by the feventh hereof); therefore in the Triangle A DE, the Right Line B C being Pa rallel to the Bafe DE, cutteth the Sides A D and A E proportionally; and therefore it follows, that
As A B is to A D :: So is AC: to A E.
(3.) Again, By the Point B let the Right Line B F be drawn Parallel to the Bafe A E, and it fhall cut the other Two Sides DA and DE proportionally in the Points B and F, (by the last hereof) and the Proportion will be
As A B AD::FE: DE; or,
For FE and BC are Equal, (by the fecond hereof). And fince it is that,
As A B AD:: AC: AE,
And fo B C to D E, they fhall alfo be,
As A CA E:: BC: D E.
For thofe Two Things, which are agreeable to a third Thing, are agreeable one to another: Therefore it generally follows,
(1.) As A B AD: BC: DE
Or, by tranfpofing the Second and Third Terms of the Proportion,
And thus it is demonftrated, that all Plain Equiangled Trian gles (as thefe, A B C and ADE are) have their Sides comprehending their Equal Angles, proportional.
IX. If feveral Plain Triangles, (how many foever) be compound. ed, and be cut by Parallel Right Lines, the Inter-Segments are Proportional. As this:
If the Two Triangles HFI and IFK be compounded, and be XXXVIII. cut with the Parallel Lines G L M, and H IK, their InterSegments are proportional. For,
As GLHI:: L'M: IK. Or,
As GL LM:: HI: IK
(by the Second and feventh hereof) For the Triangles F G L and FHI are Equiangled, (by the firft hereof) becaufe GL and HI are Parallel: Therefore,
As FL: FI:: GL: HI
(by the Seventh hereof) but (by the fame seventh hereof)
And those that are agreeable to a third, are alfo agreeable be-
X. If any one Side of a Plain Triangle be continued, the outward Angle (made by that Continuation) is equal to the Two inward and oppofite Angles of the fame Triangle.
If, in the Plain Triangle NOP, the Side N P be continued to xxxix. Q, the outward Angle OP Q fhall be equal to the Two inward Angles, ON P and NOP: For,
If from the Point P, a Right Line P R, be drawn Parallel to NO, the outward Angle O P Q fhall be compounded of the Angles OP R, and RPQ, but the Angles R PQ and R P O, are equal to the Two inward Angles ONP and PON; that is to fay, the Angle R PQ, to the Angle ON P, and the Angle OPR, to the Angle NOP, (by the firft hereof) because of the the Parallels NO and PR, and therefore the Angle OPQ is equal to the Two inward and oppofite Angles, O N P and NOP: Which was to be demonftrated.
XI. The Three Angles of every Plain (or Right Lined) Triangle, are equal to Two Right Angles..
As in the Plain Triangle N O P (in the Diagram before of the tenth) I fay, the Three Angles N OP, OPN and ON P, are XXXIX. together equal to Two Right Angles.
For the Angles (how many foever) meeting in One Point, in one and the fame Right Line, are equal to Two Right Angles.
But the Three Angles NOP, OP N and ON P, are equal to the Three Angles, meeting in the Point P, upon the fame Right Line N Q. For the Angle OP N is common to both, and the Angles RPQ and NOP (by the laft.) Therefore, the Three Angles NO P, OPN and ON P, are Equal to Two Right Angles. And from hence will follow thefe
1. That there can be but One Right, or One Obtufe, Angle, in any Plain Triangle.
2. And if One Angle be Right, or Obtufe, the other Two fhall. be Acute.
3. That the Third Angle of any Plain Triangle is the Complement of the other Two to Two Right Angles.
4. That if Two Triangles be Equiangled in any Two of their Angles, they are wholly Equiangled.
XII. In every Right-angled Plain Triangle, The Square made of the Side which fubtendeth (or is oppofite to) the Right Angle, hall be Equal to both the Squares, which are made of the Two Sides which fubtend the Right Angle. Eucl. Lib. 1. Pr. 47.
As in the Plain Right-angled Triangle S T V, Right-angled at Fig. XL. T, Ifay, the Sides ST and T V, including the Right Angle ST V,
are equal in Power to the Hypotenufe V S; that is, the Squares of the Sides S T and T V; namely, the Squares S W T E, and VTB C, added together, are equal to the Square of the Hypotenufe VS; to wit, the Square XYSV.
For, if from the Right Angle at T, be let fall the Perpendicu lar TAZ, then out of the Square XY VS is made Two Redangled Figures, ASY Z and AVX Z, the Rectangle (or Oblong) ASYZ equal to the Square SWT, and the Rectangle X ZVA, equal to the Square V TBC.And therefore the Square VXY Z, compounded of thofe Two Oblongs, is equal to the Two Squares VTBC and STEW.
But that the Two Oblongs AZ Y S and A V X Z are equal to the Squares SWT Æ and V TBC, fhall thus be proved. If Three Right Lines be Proportional, (as by the fixth hereof) the Square of the Means is equal to the Oblong made of the Two Extreams.
But the Three Right Lines S Y, (equal to S V) ST and S A, are Proportional; that is,
As S Y (SV): ST: ST: SA:
Therefore the Square of S T is equal to the Oblong made of SY (SV) and S A; for the Triangles ST V and TAS are Equiangled, becaufe of the common Angle at S, and the Two Right Angles at T and A: Therefore (by the eighth hereof)
In the fame manner it is alfo proved, that the Oblong VXZA is equal to the Square VTBC: For the Triangles TSA and ATV are Equiangled, because of their common Angle at V, and the Two Right Angles at T and A. Therefore,
As S V TV:: TV: A V.
And fo (by the fixth hereof) the Square of TV is equal to the Oblong ZX A V. Which was to be demonftrated.
Hence it followeth: That,
If in a Right-angled Plain Triangle, any Two Sides be given, the
As if the Two Sides including the Right Angle TV 8, and
XIII. In every Plain Triangle (as well Right as Oblique angled) the Sides are in Proportion one to the other, as are the Sines of the Angles oppofite to thofe Sides, & contra.
Let the Triangle A B C, be infcribed in a Circle, and from the Fig. XLI. Centre D, draw the Radii D F, DE, DG, bifecting as well the Peripheries as the Subtenfes; and let there be alfo drawn the Radius D C. Now, because the Angle at the Centre EDC is equal to the Angle, in the Peripherie A B C, and CDF, equal to CAB (by the twentieth of the third of Eucl.) Therefore fhall the Halves of the Sides be as Sines; and what Proportion the Side C A hath to the Side C B, the fame fhall the Sine HC, have to the Sine CI: For what Proportion the Whole hath to the Whole, the fame fhall the Half have to the Half. Which was to be demonftrated.
And from hence follow these
I. If the Angles of a Triangle be given, the Reason of the
II. If One Side be given, befides the Angles, both the other Sides
are alfo given.