also at C and E, be Equal one to another, each to its Correspondent : I say, the Sides about the Equal Angles are Proportional: For,

:

1. As A B: is to BC :: So is AD: to DE.
2. As A B: is to AC :: So is AD: to D E.
3. As AC: is to CB :: So is AE: to ED.

DEMONSTRATION.

(2.) Because the Angles BAC and DAE are equal, therefore if A B be applied to A D, then A C shall of Neceffity fall in AE; and by fuch Application shall such a Figure be made.

In which Figure, because that A B and AC do meet together, and also the Angles at Band D, and at Cand E, are Equal; therefore the other Sides BC and D E shall be Parallel, (by the first bereof). But in a plain Triangle a Right Line, Parallel to the Bafe, cutteth the Sides proportionally (by the seventh hereof); therefore in the Triangle ADE, the Right Line BC being Parallel to the Base DE, cutteth the Sides AD and A E proportionally; and therefore it follows, that

As AB: is to AD :: So is AC: to A E.

(3.) Again, By the Point B let the Right Line BF be drawn Parallel to the Base A E, and it shall cut the other Two Sides DA and DE proportionally in the Points Band F, (by the last hereof) and the Proportion will be

As AB:AD::FE:DE; or,
AB:AD::BC:D E.

:

For FE and BC are Equal, (by the second hereof). And fince it is that,

As AB:AD::AC:AE,

And fo BC to DE, they shall also be,

As AC:AE::BC: DE.

For those Two Things, which are agreeable to a third Thing, are agreeable one to another: Therefore it generally follows,

Fig.

Or, by transposing the Second and Third Terms of the Proportion,

thus;

:

(1.) As AB: BC::ADDE
(2.) As AB: AC::AD:AE
(3.) As AB: BC::AE:DE

And thus it is demonftrated, that all Plain Equiangled Triangles (as these, ABC and ADE are) have their Sides comprehending their Equal Angles, proportional.

IX. If several Plain Triangles, (how many foever) be compounded, and be cut by Parallel Right Lines, the Inter-Segments are Proportional. As thus:

If the Two Triangles HFI and IFK be compounded, and be XXXVIII. cut with the Parallel Lines G LM, and H IK, their InterSegments are proportional. For,

As GL:HI::LM:IK. Or,
As GL:LM::HI:IK

(by the Second and seventh hereof) For the Triangles FGLand
FHI are Equiangled, (by the first hereof) because GLand HI
are Parallel: Therefore,

As FL:FI::GL:HI

(by the seventh hereof) but (by the same seventh hereof)
As FL: FILMI Κ.

And those that are agreeable to a third, are also agreeable be-
tween themselves. Therefore,

As GL:HI::LM:IK.

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 opposite Angles of the fame Triangle.

Fig:

If, in the Plain Triangle NOP, the Side NPbe continued to xxxix. Q, the outward Angle OPQ shall be equal to the Two inward Angles, ONP and NOP: For,

If

If from the Point P, a Right Line P R, be drawn Parallel to NO, the outward Angle OPQ shall be compounded of the Angles OPR, and RPQ; but the Angles RPQ and RPO, are equal to the Two inward Angles ONP and PON; that is to say, the Angle RPQ, to the Angle ON P, and the Angle OPR, to the Angle NOP, (by the first hereof) because of the the Parallels NO and PR; and therefore the Angle OPQ is equal to the Two inward and opposite Angles, ONP and NOP: Which was to be demonstrated.

XI. The Three Angles of every Plain (or Riglit Lined) Triangle, are equal to Two Right Angles.

23

As in the Plain Triangle NOP (in the Diagram before of the Fig. tenth) I say, the Three Angles NOP, OPN and ONP, 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, OPN and ONP, are equal to the Three Angles, meeting in the Point P, upon the fame Right Line N Q. For the Angle OPN is common to both, and the Angles RPQ and NOP (by the laft.) Therefore, the Three Angles NOP, OPN and ON P, are Equal to Two Right Angles. And from hence will follow these

COROLLARIES.

1. That there can be but One Right, or One Obtuse, Angle, in any Plain Triangle.

2. And if One Angle be Right, or Obtufe, the other Two shall 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 subtendeth (or is opposite to) the Right Angle, shall 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 STV, Right-angled at Fig. XL. T; I say, the Sides ST and TV, including the Right Angle ST V,

are

are equal in Power to the Hypotenuse VS; that is, the Squares of the Sides ST and TV; namely, the Squares SWTÆ, and VTBC, added together, are equal to the Square of the Hyporenuse VS; to wit, the Square XYSV.

For, if from the Right Angle at T, be let fall the Perpendicular TAZ, then out of the Square XY VS is made Two Rectangled Figures, ASY Z and AVX Z, the Rectangle (or Oblong) ASYZ equal to the Square SWTE, and the Rectangle XZVA, equal to the Square VTBC. And therefore the Square VXYZ, compounded of those Two Oblongs, is equal to the Two Squares VTBC and STEW.

But that the Two Oblongs AZYS and AVXZ are equal to
the Squares SWT and VTBC, shall 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 SY, (equal to SV) ST and SA, are
Proportional; that is,

1

As SY (SV):ST::ST:SA:

Therefore the Square of ST is equal to the Oblong made of
SY(=SV) and SA; for the Triangles STV and TAS
are Equiangled, because of the common Angle at S, and the
Two Right Angles at T and A: Therefore (by the eighth hereof)

As SV:ST::ST:SA.

In the fame manner it is also 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 SV:TV::TV: AV.

And fo (by the fixth hereof) the Square of TV is equal to the
Oblong ZX AV. Which was to be demonstrated.

Hence

Hence it followeth: That,

If in a Right-angled Plain Triangle, any Two Sides be given, the
Third may be said to be given also.

As if the Two Sides including the Right Angle TV 8, and Fig. XL. TS 6 were given, their Squares 64 and 36 added together, make 100; the Square Root whereof is 10, for the Third Side (or Hypotenuse) SV- On the contrary, if the Hypotenuse SV 10, and one of the containing Sides T S 6, be given; substraft the Square of 6, viz. 36, from the Square of 10, viz. 100, the Remainder will be 64; the Square Root whereof is 8, for the other containing Side T V.

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 opposite to those Sides, & contra.

DEMONSTRATION.

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; bisecting as well the Peripheries as the Subtenses; and let there be also drawn the Radius DC. 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 shall the Halves of the Sides be as Sines; and what Proportion the Side C A hath to the Side C B, the same shall the Sine HC, have to the Sine CI: For what Proportion the Whole hath to the Whole, the same shall the Half have to the Half. Which was to be demonstrated.

And from hence follow these
CONSECTARIES.

I. If the Angles of a Triangle be given, the Reason of the
Sides is also given.

II. If One Side be given, besides the Angles, both the other Sides
are also given.