PROP. VII. As the base of any plane triangle ABC, is to the fum of the two sides, so is the fire of half the vertical angle, to the ca-fine of half the difference of the angles at the base. F In AC, produced, take CD = CB; join B, D, and draw CE E parallel to AB, and CF perpendicular to BD. Since CD - CB, A B D+ CBD (by 9. 1.) Moreover, seeing DCB is = the sum of the.. angles A and CBA, at the base (by 9. 1.) it is evident that BCF (or DCF) is equal to half that sum; and, therefore, as ECF is the excess of the greater ABC = BCE (by 7. 1.) above the half sum (BCF), it must, manifestly, be equal to half the difference of the same angles A and CBA. But (by Theor. 3.) AB: AD (AC+ BC):: fine D(ACB): sine ABD = sine CED (by Cor. 1. to. 7. 1.) = fine FEC = co-line ECF. 2: E. D. PROP. VII, As the base of any plane triangle 'ABC, is to the difference of the two sides, so is the co-fine of balf the vertical angle, to the line of half the difference of the angles at the base. D In the greater side CA let there be taken CD = CB, and let BD be drawn, and likewise, ce, per pendicular to BD. It is LA manifest, because CD = CB, that CDB and CBD are equal to one another, and that each of them is also equal to half the sum of the angles CBA and A at the base (by Cor. 2. to io, 1.); therefore ABD, being the excess of the greater CBA above the half fum, must consequently be equal to half the difference of the fame angles. But (by Theor. 3.) AB : AD (AC-BĆ):: sine D (co-fine DCE, or {C): sine ABD. 2. E. D. PROP. IX. As the difference of the two sides. AC, BC, of a plane triangle, is to the difference of the segments of the base AQ, BQ (made by letting fall a perpendicular from the vertex), fo is the fine of half the vertical angle, to the co-fine of half the difference of the angles at the baje. For, AC-BC:AQ BQ : : AB : AC + BC (by 9. 2. and 10.4.)::sine ACB BA : co-fine B (by Prop. 7.) Q. E. D. PROP. X. As the sum of the two sides of a plane triangle, is to the difference of the segment of the base (see the preceding figure), fo is the co-fine of balf the vertical angle, to the fine of half the difference of the angles at the base. For, 2 2 * For, AC+BC: AQ-BQ:: AB : AC-BC ACB (by 9. 2. and 10. 4:) :: co-fine of .: sine of B A (by Prop. 8.) 2: E. D.. 2 2 F As the tangent of the vertical angle C of a plane triangle ABC, is to radius, so is balf the base AB to a fourth proportional; and as half the base is to the excess of the perpendicular above the said fourthproportional, so is the fine of the vertical angle, to the co-fine of the difference of the angles at the base. Let ABCD be a circle described about the triangle, and from o, the center thereof, let OB and OC be drawn; moreover, draw CD parallel to BA, A meeting the periphery in D, and EOF, perpendicular, to AB, meeting DC in E. Then it is evident, that EF will be equal to the perpendicular height of the triangle, EOB equal to the vertical angle ACB, and FOC (=DAC) equal to the difference of the angles (ABC and BAC) at the base. But (by Theor. 2.) as tang. EOB (ACB) : radius :: EB ({AB): EO; moreover, as EB : OF (EF - EO) :: sine EOB (ACB): sine OCF, or co-line of FOC. 2. E. D. E PROP. PROP. XII. vi As the tangent of the vertical angle, of a plane triangle ABC, is to rådiús, so is the balê AB to a fourib-proportional ; and, as the said fourth-proportional, is to the sum of lhe semi-base and the line CD bifecting the base, sozis the difference of these two, to the perpendicular beigst of the triangle. D Let a circle be described also let CF, parallel to AB, B be drawn, meeting. DO, produced (if need be) in F. It is evident that DF will be pèrpendicular to AB, and equal to the height of the triangle. But DC = OC? + OD' + 200 x OF (by 11. 2.) =OA'+OD?+ 2ODxDF --UD - OA’ - OD' + 2OD X DF = AD + 2OD * DF ( by 7.2.);, whence, by taking away ADfrom the first and last of these equal quantities, we have DCP-AD:- 20DXDF; orDC+ADXDC-AD 320D XDF (by 7. 2.) and therefore 20D: DC + AÐ : : DC AD : DF; but (by Theor. 2.) as the tang. AOD=ACB (by 10.3.): to radius AD: OD): ; AB : 20D. 2. E. Di PROP. XIII. As twice the text angle under the base and perfend'icular of a plane triangle ABC, is to the rectangle under the suit; and difference, of the base and sum of the two fides" ; jo is radius, to the co-tangent of half tl-eozertical engle. Let 1 D E B В 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 triangle; also let cf be A parallel to AB, and let HA, HB and HC, be H drawn. Since the diameter HG is perpendicular to AB, therefore is AE = BE (by 2. 3.) AH = BH, and the angle ACH=BCH (by 1 2. 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 manifest, that 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 triangles AEH and HCD, it will be, AE:.CD2 :: HA? -HEX HG (by Cor. to 19.4.): HC" (HFXHG) :: HE : HF (by 7. 4.) Whence, by division, &c. AE : CD'-AE:: HE: EF :: HEXAE: EF XAE; therefore, by inversion and alternation, EF XAE : CD-AE(CD+AEX CD-AE):: HEXAE:AE, HÈ: AE ::radius : Co-tang. EAH (ACH, by Theor. 2.): whence the truth of the proposition is manifeft. PROP. XIV. As twice the rectangle of the base and perpendicular of a plane triangle ABC, is to the rectangle under the sum, and difference, of the base and the difference of the two sides; fo is radius, to the tangent of half the vertical angle. F Let |