Cafe 2. Let C be an obtuse angle, so that the perpendicular DB falls without the triangle. Then, because AC2+ CB2+ 2AC.CD-AB2 (12. 2.), or AC2+CE2+ 2AC.CD=AB2; and because AČ2+CÉ2- 2AC.CE+AE2, 2AC.CE+ 2AC.CD+ AE2= AB2. Now,

=

2

2AC.CE+2AC.CD= 2AC.ED, therefore 2AC.ED+AE AB2. Therefore, it will be demonftrated as before, that 4AC.CE: (AB+AE)(AB-AE) :: R2: (fin. + C)2.

B

E

Now, 4AC.CE, or 4AC.CB is four times the rectangle under the fides, and AE being the difference of the fides, (AB+AE)(AB-AE) is the rectangle which has for one of its fides the base increased by the difference of the two fides of the triangle, and for its other fide, the bafe diminished by the fame difference. Therefore, in any triangle, &c. Q.E. D.

COR. Because BDC is a right angle, BC: CD:: R: fin.CBD; now, fin. CBD cof. C, therefore, BC: CD:: R: cof. C, and 2AC.CB: 2AC.CD :: R: cof. C. But, (13.2.) 2AC.CD=AC2+CB2-AB2, therefore,

2AC.CB: AC2+CB2-AB2 :: R: cof. C; that is, twice the rectangle AC.CB, is to the excefs of the fquares of AC and CB above the fquare of AB, as radius to the co-fine of the included angle ACB.

PROP

A

'PRO P. VIII.

OUR times the rectangle contained by any two fides of a triangle, is to the rectangle contained by two ftraight lines, of which one is the fum of those fides increased by the bafe of the triangle, and the other the fum of the fame fides diminished by the base, as the fquare of the radius to the fquare of the co-fine of half the angle included between the two fides of the triangle.

The fame conftruction being made as in the last propofition, it is plain, that AF is the fum, and AE the difference of the fides AC and CB, and also, that the angle FEB is half of the angle FCB. Now, if ACB be an acute angle AC2+CB2, or AC2+CF2=AB2+ 2AC.CD; add 2AC.CF to both, and AC2+CF2+ 2AC.CF, that is, AF-AB2+2AC.CD+2AC.CF; But

2AC.CD+ 2AC.CF= 2AC.DF, because CD+CF=DF, and therefore, AF-AB2+ 2AC.DF; and

2AC.DF-AF2-AB2= (AF+AB)(AF-AB), (2. 5.). Again, AC CF :: 2AC.DF: 2CF.DF, and

:

4AC.CF: 4CF2 :: 2AC.DF: 2CF.DF; and alternately, 4AC.CF: 2AC.DF:: 4CF2: 2CF.DF; now,

4CF-EF2, and 2CF.DF-EF.DF-FB2, therefore,

4AC.CF: 2AC.DF:: EF2 : FB2. Bnt

EF: FB:: R: fin. FEB, and fin. FEB cof. EFB, because EFB is the complement of FEB, so that

fin. FEB cof. ECB; therefore, EF: FB:: R: cof. LECB,

2

and EF2: FB2 :: R2: (cof. ECB.)

4AC.CF: 2AC.DF:: R2: (cof.

2

Therefore,

ECB)2, or

4AC.CB: (AF+AB)(AF—AB): : R2: (cof. 4 ECB).2

2

In the fame way, when the angle C is obtufe, it will be demonftrated, that 4AC.CB: (AF+AB)(AF—AB) : : R2 : (cof. ECB). Now, 4AC.CB is four times the rectangle under the fides, and AF being the fum of the fides, (AF+AB)(AF-AB) is the rectangle which has one of its

fides equal to the fum of the fides of the triangle increased by the base, and its other fide equal to the fum of the fides of the triangle diminished by the bafe. Q. E. D.

COR. Because 4AC.CB: (AF÷AB) (AF—AB) : : R2 : (cof. ECB)2, and also (prop. 7.) 4AC.CB : (AB+AE)(AB-AE):: R2 (fin. ECB); therefore, ex æquo, (AF+AB) (AF-AB): (AB+AE)(AB-AE):: (cof. ECB)2: (fin. † ECB)2. 50 2

But, cof. ECB: fin. ECB:: R: tan. 1 ECB;

wherefore,(AF÷AB)(AF—AB): (AB+AE)(AB—AE):: R2: (tan. ECB)2; that is, the rectangle contained by two ftraight lines, of which one is the fum of the three fides of a triangle, and the other the fum of two of the fides of the triangle diminished by the third, is to the rectangle contained by two ftraight lines, of which one is the third fide of the triangle increased by the difference of the other two, and the other the third fide diminished by the difference of the other two, as the fquare of the radius to the fquare of the tangent of half the angle included between the two fides of the tri angle.

LEMMA II.

F there be two unequal magnitudes, half their

I difference added to half their fun is equal to

the greater; and half their difference taken from half their fum is equal to the lefs.

Let AB and BC be two unequal magnitudes, of which AB is the greater; fuppofe AC bifected in D, and AE equal to BC. It is manifeft, that AG is the

fum, and EB the difference A E D B

of the magnitudes. And be

cause AC is bifected in D, AD is equal to DC; but AE is alfo equal to BC, therefore DE is equal to DB, and DE or DB is half the difference of the magnitudes. But AB is equal to BD and DA, that is to half the difference added to half the fum; and BC is equal to the excefs of DC, half the fum, above DB, half the difference. Therefore, &c. Q.E. D. COR. Hence, if the fum and the difference of two magni. tudes be given, the magnitudes themselves may be found.

SOLUTION of the Cafes of Right Angled
Triangles.

IN

PROBLEM.

N a right angled triangle of the three fides and three angles, any two being given, befides the right angle, and one of those two being a fide, it is required to find the other three.

The reafon of the limitation in this Prop. to those cafes in which at least one fide is given, is manifeft; for, if the angles only be given, the fides cannot be found, but only their ra tios, which are the fame with the ratios of the fines of the oppofite angles,

It is evident, that when one of the acute angles of a right angled triangle is given, the other is given, being the complement of the former to a right angle; and also that the fine any of the acute angles is the co-fine of the other.

of

As this general problem admits of several cases, the folutions or rules for calculation, which all depend on the first Prop. may be conveniently exhibited in the form of a table.

When two fides of a right angled triangle are given in numbers, the third may be found, either trigonometrically, as in the table, or by the 47th of the 1ft of the Elem. Thus, becaufe BC AB2+AC2, if AB and AC are given, BC is found, for it is the square root of AB+AC2. In the fame manner, if AB and BC are given, AC is found, being the fquare root of BC2— AB2.