difference is to the distance of a perpendicular from the middle of the base. This distance, added to half the base, gives the greater segment, and deducted from half the base, gives the less.

Let the proposed triangle be ABC; from the centre

C, with CB, the shortest side, as radius, describe the circle FGBH; produce AC to H. Because CB= CH, AH=AC+CB, the sum of the sides; and as CB=CF, AF-AC-BC, the difference between the sides. Again, the perpendicular CD bisects GB (3.3); hence AG is evidently the difference between the segments AD and DB, and ED being half the difference between AD and DB, it follows that ED= AG. Finally, AH× AF-AB× AG (36.3). Hence (16.6), AB : AH::AF: AG; that is, AB: AC+ CB:: AC-CB : 2 ED.

Now, if the first term in this proportion be multiplied by 2, and the fourth divided by 2, these, with the two middle terms, will form a proportion. Hence 2 AB: AC+CB:: AC-CB: ED; then ED added to AE, gives the greater segment AD, and ED being deducted from EB, leaves the less segment DB.

D

COR. 1. Hence, the base, or longest side, is to the sum of the other two, as their difference is to the difference of the segments of the base, made by a perpendicular demited upon the longest side, from the opposite angle; then from half the base deduct half this difference, for the less segment, and to half the base add the half difference, for the greater segment.

To shew this, let sr and rm be any two unequal

quantities, of which sr is the greater. Make sd=rm, and dr is the difference between sr and rm; bisecting dr in b, db or br is half the difference between the segments. Now, as sd=rm, and db-br, sb=bm; and if to sb (=3m), br (=) be added, the sum sr will be the greater segment; and if from bm (=TMm ), br be deducted, the remainder rm will be the less seg

ment.

sm

2

COR. 2. If from the greater of two unequal quantities half their sum be taken, the remainder will be half their difference.

6. In any plane triangle, the rectangle contained by any two sides is to the rectangle

C

bisect the exterior angles made by these with the base AB, by the lines BP and AP, meeting at P; draw the perpendicular PF, and make AE AF, and BD=BF; join PE and PD; having joined CP, upon it let fall the perpendiculars AG and BH.

In the triangles APF and APE, the sides AE and AF are equal, and AP is common; also the angles EAP and FAP are equal: therefore the sides EP and PF are equal, and also the angle at E and F are equal; but the angle at F is right; then the angle at E is right. For the same reason, FP-PD, and the angle at D is right: hence PE-PD. And as APE and APF are in every respect equal, as likewise FPB and DPB, the angle APB is half of the angle EPD. In the triangles EPC and DPC, EP2+EC2=PC2= PD2+DC2; but EP2-PD2; therefore EC-DC2: hence EC=DC, and PC being common to both the the triangles, then the angle ECP=DCP, and EPC =DPC. Now, as AE and BD are together equal to AB, it follows that CE+CD=AC+CB+AB, the perimeter, and CE or CD is equal to half the perimeter: AE and BD are evidently the excesses of half the perimeter above the sides AC and AB.

As the angle APB is equal to half the angle EPD, it is equal to each of the angles EPC and CPD; take from these equals the common part APC, and there remains BPC-EPA; and taking away CPB, there remains APC BPD. Hence the two right-angled triangles PAG and PBD, and the two triangles PBH and APE are respectively equiangular; then, by 4.6, we have

D 2

52

ELEMENTS OF PLANE TRIGONOMETRY.

AG: AP BD: BP, and

AP: AE :: BP :: BH; then ex. æquali,
AG: AE:: BD: BH; hence, (16.6)
AGXBH AE×BD.

Again, in the right-angled triangles AGC and BCH,
AC: AG: R : sin C (Prop. 1.)

BC: BH: R sin C (Prop. 1.)

Hence, ACX BC: AG×BH=AE×BD:: R2 : sin2C, which is the expression for the proposition.

53

SECTION IV.

On the Solution of Plane Triangles.

In the solution of plane triangles, it is necessary to attend to two distinct processes.

The first is to establish general formulæ, showing in every case the relation of the required parts of any triangle to those which are given, so that the former may be computed from the latter. This was the subject of the last chapter.

The second is the calculation of certain numerical tables, in which are registered the sines, cosines, tangents, &c. or their logarithms; by means of which, the required parts of any plane triangle may be computed, when the general formulæ before mentioned are properly applied.

This subject, however, requiring too extensive a course of purely elementary science, and wishing to introduce the practice at as early a stage as possible, the method of computing the tables is reserved for a subsequent work.

Though the student is referred to a subsequent work for the actual computation of the tables, yet it is necessary in this place to attend to the nature of them, in order to assist him in the determination of certain formulæ, when particular numbers are substituted for their general symbols.

In the preliminary observations prefixed to the last