Let the preceding figure and conftruction be re tained, 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 AD: AE:: GC' (GFX HG): AG' (GE× HG): : GF : GE; therefore, by divifion, &c. AE-AD': AE:: EF: GE:: EFXAE: GEXAE; whence, again, by alternation, &c. EFX AE: AE-AD2 (AE+AD × AE-AD):: GE× AE: AE2::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 a: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,

C

Make AB equal to a, upon which let a femicircle ADB be defcribed; 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. Because ACXCB=CD1b2 (by Cor. to 19. 4.) it is plain that ACX-AC, or BC-a-BC is also=b2; and, therefore, xxa-x being b', it is manifest that x may be equal, either, to AC, or to BC. Now

(by

(by Theor. 1.) OD (a): 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): AC. Q. E. Ď.

PROP. XVI.

D.

If the relation of three lines, a, b and x, be such, that xax 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 fign of ax is pofitivè or negative) : : b: x.

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

F

E

A

joined, and draw BF parallel to AC, meeting AE, produced, in F. Then, fince (by 22. 3.) BEX BD (=BE × BE + a = BD × BD —− a) =AB2 (b2). and xxx+ab, by fuppofition, it is manifeft, that BE will be x, when xxx+ab; and BDx, when xxx—a—b1.

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 (6) F 2 : BF

: BF (BE), the value of x in the first cafe, where x2 + 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 fecond cafe; where x-ax b3. Q. E. D.

PROP. XVII.

In any plane triangle ABC, it will be, as the line CE bifelting the vertical angle, is to the bafe AB, jo is the fecant of half the vertical angle A B, 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

Let ACBD be a circle defcribed about the triangle, and let CE be produced to meet the periphery thereof in D; moreover, let AD B 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 DBFACD (by 12. 3.) DCB, the triangles DEB and DBC (having D common) are equiangular, and therefore DEX DC = DB2 (by 24. 3.) or, which is the fame, DE + DE×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.

www

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

first

first proportion, we have, CEXDB: AB × 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 balf 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.

F

p

E

B

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: BE':; DC2 (DF × DP): BD2 (DE×DP) :: DF (DE+EF); DE :: BE+ EI: EB:: BE+ EIxBE: EB2; and, confequently, CG2 (=BE + EI × BE) = BE2+ EI 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

X

(::

(: : 2CG, or AC + BC, to AB). But (by Theor. 2.) EF: EI :: tang. I. (ACD) : radius; therefore, by compounding this proportion with the laft but one, we fhall have, EI x EF: EI XCG:: tang. ACD xradius: tang. Qx radius (by 11. 4.) and confequently EF: 2CG (AC+ BC) :: tang. ACD: tang. Q. Whence the truth of the propofition is manifeft.

PROP. XIX.

In any plane triangle ABC, it will be, as the perpendicular is to the difference of the two fides; fo is the co-tangent of half the vertical angle, to the tangent of an angle; and, as radins is to the co-tangent of half this angle, fo is the difference of the fides to the bafe of the triangle.

E

Α

P

F

D

Let DP, DG, CF, &c. be as in the preceding propofition; alfo let PA and PC be drawn, and FI, pa rallel to PA, meeting AB in I.

The right-angled triangles ADG and DPC, having DAG = DPC (by Ccr. to 12. 3.) are fimilar;

and therefore, AG: PC:: AD: DP :: AE'; AP (by Cor. to 11. 4.); whence, alternately, AG AE : PC (PF x PD): AP2 (PE × PD):: PF: PE:: AI: AE:: AI × AE: AE2 ·(by 7.4.); and confequently, AG' = AI × AE= AE-EIX AE. Therefore, by Prop. 16, EI: AG : radius; tangent of an angle (Q); and as radius: co-tang. Q::AG: AE. But (by Theor. 2.) EF EI co-tang. EFI (ACD): radius; which proportion being compounded with the last

3

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