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 fides of the triangle; alfo let CF be A parallel to AB, and let

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, likewife, AD = Bd (by 15. 1.) From whence it is manifeft, that CD will be equal to half the fum, and AD equal to half the difference, of the two fides of the triangle. Moreover, because of the fimilar triangles AEH and HCD, it will be, AE: CD':: HA' =HEX HG (by Cor. to 19. 4.): HC2 (HF × HG) :: HE: HF (by 7. 4.) Whence, by divifion, &c. AE: CD-AE:: HE: EF:: HEX AE: EF x AE; therefore, by inverfion and alternation, EF × AE : CD2 - AE2 (CD + AE × CD-AE) :: HEX AE AE. HE AE:: radius: co-tang. EAH (ACH, by Theor. 2.): whence the truth of the propofition is manifeft.

PROP. XIV.

As twice the rectangle of the base and perpendicular of a plane triangle ABC, is to the rectangle under the fum, and difference, of the bafe and the difference of the two fides; fo is radius, to the tangent of half the vertical angte.

Let the preceding figure and conftruction be retened, and let AG and CG be drawn. The triangles AHD and GHC (being right-angled at D and C, and having HAD = HGC (by 11. 3.) are equiangular; and fo AD: GC:: AH: HG:: AE: AG (by 19. 4.); whence, alternately, AD: AE :: GC: AG, and AD2 : AE2 :: GC2 (GF × HG): AG' (GE × HG) :: GF: GE; therefore, by divifion, &c. AE - AD: AE2:: EF: GE:: EFX AE: GE X AE; whence, again, by alternation, &c. EFX AE: AE-AD2 (AE+AD × AE AD):: GEX AE: AE:: GE: AE :: radius: tang. AGE (by Theor. 2.); from which the truth of the propofition is manifeft.

PROP. XV.

If the relation of three right-lines a, b and x, be fuch, that ax-x2= b2; then it will be, as ža: b :: radius to the fine of an angle; and, as radius, to the tangent (or co-tangent) of half this angle, fo is b: x.

A

D

C

Make AB equal to a, upon which let a femicircle ADB be described; B alfo let CD, equal to b, be perpendicular to AB, and meet the periphery in D (for it cannot exceed the radius of the circle when the propofition is poffible): moreover, let AD, BD, and the radius OD, be drawn. Becaufe ACX CB = CD2= b2 (by Cor. to 19. 4.) it is plain that AC xa-AC, or BC x a- BC is also = b2; and, therefore, xx-x being a x being = b2, it is manifeft that x may be equal, either, to AC, or to BC. Now

2

(by

(by Theor. 1.) OD (ža) : CD (b) :: radius: fine DOC; whofe half is equal to A, or BDC (by 10. 3.) But, as radius: tang. BDC:: DC (b): BC; or, as radius co-tang. BDC (tang. CDA) :: DC (b) : AÇ. 2. E. D.

PROP. XVI.

If the relation of three lines, a, b and x, be fuch, that x2+ax = b2; then it will be, as ža : 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 fign of axis pofitive or negative): :b: x.

Make AB = b, and AC, perpendicular to AB, equal to a; about the latter of which, as a diameter, let a circle be defcribed; and, thro' O, the center thereof, let BD be drawn, meeting the periphery in E and D; also let A, E and C, E be

D

F

B

E

A

joined, and draw BF parallel to AC, meeting AE, produced, in F. Then, fince (by 22. 3.) BEX BD (= BE X BE+a= BD x BD — a) — AB2 (b2) and x x x + a = b', by fuppofition, it is manifeft, that BE will be = x, when x x x + a = b2; and BDx, when xxx a = b2.

Furthermore, because the angle FOAE (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 likewife equal.

Now (by Theor. 2.) AO (a: AB (b): radius: tang. AOB; whofe 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 cafe, where x2 + ax = b2. 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.); 10.4.); whence, by equality, radius: co-tang. BAF :: AB (b): BD; which is the value of x in the second cafe; where — ax = b2. Q, E. D. x2

PROP. XVII.

In any plane triangle ABC, it will be, as the line CE bifecting the vertical angle, is to the base AB, so is the fecant of half the vertical angle ACB, to the tangent of an angle; and, as the tangent of half this angle is to radius, fo is the fine of half the vertical angle, to the fine of either angle, which the bifelting line makes with the bafe.

FE

A

D

B

Let ACBD be a circle described about the triangle, and let CE be produced to meet the periphery thereof in D; moreover, let AD and BD be drawn, and likewife DF, perpendicular to the bafe AB; which will, alfo, bifect it, because (BCD being ACD) the fubtenfes BD and AD are equal (by 10. 3.) Moreover, fince the angle DBE = ACD (by 12. 3.) = DCB, the triangles DEB and DBC (having D common) are equiangular, and therefore DE x DC = DB2 (by 24. 3.) or, which is the fame, DE + DE x CE = DB'. 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 (AB :: fecant FBD (BCE): radius; therefore, by compounding this, with the

firft

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

PROP. XVIII.

In any plane triangle ABC, it will be, as the perpendicular is to the fum of the two fides, fo is the tangent of half the angle at the vertex, to the tangent of an angle; and, as radius is to the tangent of half this angle, fo is the fum of the two fides, to the bafe of the triangle.

Let DP, perpendicular to AB, be the diameter 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 I A alfo FI, parallel to BD, meeting BA, produced, in I.

G

F

P

E

It is manifeft, from Prop. 13. that CG is equal to half the fum of the fides AC and BC: it alfo appears, (from 7. 1. and Cor. to 12. 3.) that the triangles DCG, ADE, BDE and IFE, are all equiangular. Therefore it will be, CG: BE2 :: DC2 (DF × DP) : BD2(DE × DP) : : DF (DE÷EF)': DE :: BE+EI: EB :: BE + EI × BE: EB2; and, confequently, CG2 (= BE + EI × BE) = BE2 + HIX 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

F 3

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