PROP. VII.

As the bafe of any plane triangle ABC, is to the Jum of the two fides, fo is the fine of half the vertical angle, to the co-fine of half the difference of the angles at the bafe.

In AC, produced, take CDCB; join B, D, and draw CE parallel to AB, and CF perpendicular to BD.

Since CD = CB, therefore is the angle

D DBC (by 12. 1.) and, confequently, halfthe vertical angle ACB =

= D.

D+ CBD

(by 9. 1.)

Moreover, feeing DCB is the fum of the angles A and CBA, at the bafe (by 9. 1.) it is evident that BCF (or DCF) is equal to half that fum; and, therefore, as ECF is the excefs of the greater ABC BCE (by 7. 1.) above the half fum (BCF), it must, manifeftly, be equal to half the difference of the fame angles A and CBA.

But (by Theor. 3.) AB: AD (AC+ BC): fine' D (ACB): fine ABD fine CED (by Cor. 1. to. 7. 1.) fine FEC co-fine ECF. 2. E. D. F

PROP. VII.

As the bafe of any plane triangle ABC, is to the difference of the two fides, fo is the co-fine of half the vertical angle, to the fine of half the difference of the angles at the bafe.

A

E

In the greater fide CA let there be taken CD = CB, and let BD be drawn, and likewife, CE, perpendicular to BD. It is B manifeft, because CD = CB, that CDB and CBD.

are equal to one another, and that each of them is alfo equal to half the fum of the angles CBA and A at the bafe (by Cor. 2. to 10. 1.); therefore ABD, being the excefs of the greater CBA above the half fum, muft confequently be equal to half the difference of the fame angles.

But (by Theor. 3.) AB: AD (AC-BC):: fine D (co-fine DCE, or C): fine ABD. Q. E. D.

PROP. IX.

As the difference of the two fides AC, BC, of a plane triangle, is to the difference of the fegments of the base AQ, BQ (made by letting fall a perpendi-. cular 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 bafe.

For, AC-BC: AQBQ:: AB AC + BC (by 9. 2. and 10.4.):: fine ACB

2

: co-fine

B A

2

(by Prop. 7.) Q. E. D.

PROP. X.

As the fum of the two fides of a plane triangle, is to the difference of the fegment of the bafe (Jee the preceding figure), fo is the co-fine of half the vertical angle, to the fine of half the difference of the angles at the bafe.

For,

For, AC+BC: AQ—BQ :: AB: AC-BC

(by 9. 2. and 10. 4) :: co-fine of

ACB

: fine of

2

As the tangent of the vertical angle C of a plane triangle ABC, is to radius, fo is half the bafe AB to a fourth proportional; and as half the bafe is to the excess of the perpendicular above the faid fourthproportional, fo is the fine of the vertical angle, to the co-fine of the difference of the angles at the bafe.

Let ABCD be a circle defcribed about the triangle, and from O, the center thereof, let OB and OC be drawn; moreover, draw CD parallel to BA, meeting the periphery in D, and EOF, perpendicular, to AB, meeting DC in E.

A

D

F

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 (EFEO): fine EOB (ACB): fine OCF, or co-fine of FOC. 2. E. D.

PROP.

PROP. XII.

As the tangent of the vertical angle, of a plane triangle ABC, is to radius, jo is the bafe AB to a fourth-proportional; and, as the faid fourth-proportional, is to the fum of the femi-bafe and the line CD bifecting the bafe, fe is the difference of these two, to the perpendicular height of the triangle.

F

D

B

Let a circle be defcribed about the triangle, and from O, the center thereof, let OA, OC and OD be drawn; alfo let CF, parallel to AB, be drawn, meeting DO, produced (if need be) in F. It is evident that DF will be perpendicular to AB, and equal to the height of the triangle. But DC OC' + OD' + 20D x OF (by 11. 2.) =OA'40D'+20D×DF—OD - OA - OD' + 20D x DF AD' + 20D xDF(by7.2.); whence, by taking away AD2 from the first and last of these equal quantities, we have DC-AD-20DXDF; orDC+ADXDC-AD =20D×DF (by 7. 2.) and therefore 2OD: DC + AÐ: DC-AD: DF; but (by Theor. 2.) as the tang. AOD=ACB (by 10. 3.) ; to radius (:} AD: OD):: AB: 20D. QE. D.,

PROP. XIII.

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

Let

Since the diameter HG is perpendicular to AB, therefore is AE = BE (by 2. 3.) AH BH, and the angle ACH BCH (by 12. 3.); whence, alfo, 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, 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, AE2: CD2:: HA' =HEX HG (by Cor. to 19.4.): HC (HFXHG) :: HE: HF (by 7. 4.) Whence, by divifion, &c. AE': CD-AE:: HE: EF:: HEXAE: EF XAE; therefore, by inverfion and alternation, EF × AE: CD2 — AE2 (CD+AE × CD-AE): HEXAE: 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 bafe 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 angle.