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Let HG, perpendicular to AB, be the diameter of a circle described about the triangle; and let HD and Hd be perpendicular to the two sides of the

B triangle ; also let CF be A parallel to AB, and lec HA, HB and HC be drawn.

Since the diameter AG is perpendicular to AB, therefore is AE = BE (by 2. 3.) AH = BH, and the angle ACH = BCH (by 12. 3.); whence, also, CD = Cd, and HD = Hd (by 15. 1.) Therefore, the right-angled triangles HAD and HBd, having AH = HB and HD = Hd, have, likewise, AD = Bd (by 15. 1.) From whence it is manifeft, thac CD will be equal to half the sum, and AD equal to half the difference, of the two sides of the triangle. Moreover, because of the similar criangles AEH and HCD, it will be, AE : CD: :: HA? =HE X HG (by Cor. to 19. 4.): HC(HFX HG) :: HE : HF (by 7. 4.) Whence, by division, &c. AE? : CD-AE: :: HE: EF::HEX AE: EF X AE; therefore, by inversion and alternation, EF x AE : CD – AE (CD+ AE x CD-AE):: HE X AE: AE. HE: AE :: radius : co-tang, EAH (ACH, by Theor. 2.): whence the truth of the proposition is manifeft.

PROP. XIV. As twice the restangle of the base and perpendicular of a plane triangle ABC, is to the rectangle under the fum, and difference, of the base and ibe difference of the tico fides; so is radius, in the 100gent of half the vertical ongte.

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Let the preceding figure and construction be retened, and let AG and CG be drawn. The triargies AHD and GHC (being right-angled at D and C, and having HAD = HGC (by 11. 3.) are equiangular; and so AD: GC :: AH : HG:: AE: AG (by 19. 4.); whence, alternately, AD: AE :: GC : AG, and AD: : AE? :: GC(GFX HG) : AGʻ (GE ~ HG) :: GF: GE; therefore, by division, &c. AE — AD: AE?:: EF:GE:: EF X AE:GE X AE; whence, again, by alternation, &c. EF X AE: AE - AD' (AE+ ADX AE AD) ::GE X AE: AE’::GE : AE :: ra. dius : tang. AGE (by Theor. 2.); from which the truth of the proposition is manifeft.

PROP. XV. If the relation of three right-lines a, b and x, be such, that ax - x = b’ ; iben it will be, as a:b :: radius to the fine of an angle ; and, as radius, to the tangent (or co-targent) of half this angle, se is b:x.

D

С

Make AB equal to a, upon which let a semi

circle ADB be described

B also let CD, equal to b,
A

be perpendicular to AB,
and meet the periphery
in D (for it cannot exceed

the radius of the circle
when the proposition is possible): moreover, let
AD, BD, and the radius OD, be drawn. Becaufe
AC X CB = CD'— 62 (by Cor. to 19. 4.) it is plain
that AC xã — AC, or BC Xã -- BC is also =
and, therefore, x x 2 -- * being=b, it is manifeft
that x may be equal, either, to AC, or to BC, Now

(by

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(by Tbeor. 1.) OD (2):CD (6) :: radius : sine DOC; whofe half is equal to A, or BDC (by 10. 3.) But, as radius : tang. BDC :: DC (6) : BC; or, as radius : co-tang. BDC (tang. CDA) :: DC (6) : AC. 2. E. D.

D

PROP. XVI. If the relation of three lines, a, b and x, be such, that x2 + ax = b; then it will be, as a :b :: radius to the tangent of an angle ; and as radius is to the tangent, or co-tangent, of half this angle (according as the sign of ax is positive or negative) ::b:x. Make AB=b, and AC,

B perpendicular to AB, equal

E to a; about the latter of which, as a diameter, let a circle be described ; and,

A thro' O, the center thereof, let BD be drawn, meeting the periphery in E and D; also let A, E and C, E be joined, and draw BF parallel to AC, meeting AE, produced, in F. Then, since (by 22. 3.) BE X BD . (=BE X BE ta= BD X BD-a)= AB’ (ba) and x x x + a = b, by supposition, it is manifeft, that BE will be = %; when x x x ta=b”; and BD = %, when x x * - a= b

Furthermore, because the angle F = OAE by 7. 1.) = OEA (by 12. 1.) = BEF (by 3. 1.) it is evident that BF = BE (by 18. 1.) and that the angles BAF and C (being the complements of the equal angles F and OAE) are likewise equal.

Now (by Theor. 2.) AO (:AB (b) :: radius : tang. AOB ; whose half is equal to C, or BAF (by 14. 3.) But, as radius : tang. BAF :: AB (b)

F 2

.

: BF

:BF (BE), the value of x in the first case, where *2 + ax = b. Again, radius : co-tang. BAF (tang. F) :: BF (BE): AB (by Theor. 2.);

and BE : AB :: AB : BD (by Cor. to 22. 3. and 10. 4.); whence, by equality, radius : co-tang. BAF :: AB (b) : BD; which is the value of x in the second case; where x2 - ax = b. 2. E. D.

PROP. XVII. In any plane triangle ABC, it will be, as the line CE bileeting the vertical angle, is to the base AB, so is the secant of half the vertical angle ACB, to the tangent of an angle ; and, as the tangent of half this angle is to radius, so is the fine of balf the vertical angle, to the fine of either angle, which the biseating line makes with the base.

D

Let ACBD be a circle described about the triangle, and let CE be produced to meet the periphery thereof

in D; moreover, let AD А

1 В and BD be drawn, and like

wise DF, perpendicular to the base AB; which will,

also, bisect it, because (BCD being = ACD) the fubtenses BD and AD are equal (by 10. 3.) Moreover, since the angle DBE

= ACD (by 12. 3.) = DCB, the triangles DEB and DBC (having D common) are equiangular, and therefore DE XDC = DB2 (by 24. 3.) or, which is the same, DE+- DE X CE = DB2. Therefore (by Prop. 16.) ;CE: DB :: radius': tangent of an angle (which we will call Q); and, as radius : tang. Q:: DB : DE.

But DB : BF GAB :: fecant FBD (BCE) : radius; therefore, by compounding this, with the

first

first proportion, we have, 1 CE X DB : AB x DB :: radius x fecant BCE : radius X tang. Q (by 10. 4.) and consequently CE : AB :: fecant BCE : tang. (by 7.4.) . Again, DE: DB :: fine DBE (BCE): sine of DEB (or of CEB); whence, by equality, tang. Q : radius :: line BCE : sine CEB. 2. E. D.

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PROP. XVIII. In any plane triangle ABC, it will be, as the perpendicular is to the sum of the two sides, so is the tangent of half the angle at the vertex, to the tangent of an angle; and, as radius is to the tangent of balf this angle, so is the sum of the two sides, to the base of the triangle. Let DP, perpendicular

р to AB, be the diameter

с C of a circle described about the triangle; let CF be perpendicular to DP, and DG to AC, and let DA, DC and DB be drawn, and also FI, parallel to BD, meeting BA, produced, in I.

It is manifeft, from Prop. 13. that CG is equal to half the sum of the sides AC and BC: it also appears, (from 7. 1. and Cor. to 12. 3.) that the triangles DCG, ADE, BDE and IFE, are all equiangular. Therefore it will be, CG2: BE:: DC2 (DF XDP): BD’(DE X DP):: DF (DE+EF): DE::BE +EI:EB :: BE + EI BE:EB”; and, consequently, CG? (= BE + EI X BE) = BE? + EI X BE. But, by Prop. 16. it will be, EI:CG :: radius : tangent of an angle (which we will call Q); and as radius : tang. Q:: CG: BE

(: :

B

F 3

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