therefore (by theo. 18) AB: BC::(BEA:BEC · or BEA: BED):: AE: ED. Q. E. D.

Cor. 1. Hence also AC: AB:: AD: AE: For AC: AB:: (AEC: AEB:: ABD: AEB) :: AD: AE.

Cor. 2. It also appears that a right line, which divides two sides of a triangle proportionally, must be parallel to the remaining side.

Cor. 3. Hence also, theo. 16. is manifest; since the sides of the triangles ABE, ACD, being equiangular, are proportional.

THEO. XXI.

PL. 2. fig. 4.

If two triangles ABC, ADE, have an angle BAC, in the one, equal to an angle DAE, in the other, and the sides about the equal angles, proportional; that is, AB: AD :: AC : AE; then the triangles will be mutually equiangular.

In AB take Ad = AD, and let de be parallel to BC, meeting AC in e.

Because (by the first cor. to the foregoing theo.) AB: Ad (or AD):: AC: Ae, and (by the hypothesis, or what is given in the theorem) AB: AD: AC AE; therefore Ae = AE seeing AC bears the same proportion to each; and (by theo. 6.) the triangle Ade = ADE, therefore the angle Ade D and Aed E, but since ed and BC are parallel (by part 3. theo. 3) Ade B, and AedC, therefore B-D and C-E.

=

THEO. XXII.

PL. 2. fig. 5.

Q. E. D.

=

Equiangular triangles ABC, DEF, are to one another in

a duplicate proportion of their homologous or like sides; or as the squares AK, and DM of their homologous sides.

Let the perpendiculars CG and FH be drawn as well as the diagonals BI and EL.

The perpendiculars make the triangles ACG and DFH equiangular, and therefore similar (by theo. 16.) for because the angle CAG=FDH, and the right angle AGC=DHF, the remaining angle ACG=DFH, (by cor. 2. theo. 5.)

Therefore GC: FH:: (AC: DF::)AB: DE, or which is the same thing, GC: AB:: FH: DE for FH multiplied by AB = AB multiplied by FH.

By theo. 19. ABC: ABI:: (CG: AI or AB as before FH DE or DL : :) DFE : DLE, therefore ABC: ABI:: DFE: DLE, or ABC: AK:: DFE: DM, for AK is double the triangle ABI, and DM double the triangle DEL, by cor. 2. theo. 12. Q. E. D.

THEO. XXIII.

PL. 2. fig. 6.

Like polygons ABCDE, a b c d e, are in a duplicate proportion to that of the sides AB, a b, which are between the equal angles A and B and a and b, or as the squares of the sides AB, ab.

Draw AD, AC, ad, ac.

By the hypothesis AB ab :: BC: bc, and thereby also the angle B =b; therefore (by theo. 21.) BAC = bac; and ACB a cb in like manner EAD-e a d, and EDA=eda. Iftherefore from the equal angles A, and a, we take the equal ones

=

EAD+BAC=e a d,+ b a c the remaining angle DAC da c, and if from the equal angles D and d, EDA=eda, be taken, we shall have ADC=a d c and in like manner if from C and c be taken BCA, bca, we shall have ACD a cd; and so the respective angles in every triangle, will be equal to those in the other.

=

By theo. 22. ABC: abc the square of AC to the square of ac, and also ADC: ade: the square of AC, to the square of ac; therefore from equality of proportions ABC: abc:: ADC: a dc; in like manner we may shew that ADC: a dc: EAD: ead: Therefore it will be as one antecedent is to one consequent, so are all the antecedents to all the consequents. That is, ABC is to a b c as the sum of the three triangles in the first polygon, is to the sum of those in the last. Or ABC will be to a b c, as polygon to polygon.

The proportion of ABC to a b c (by the foregoing theo.) is as the square of AB is to the square of a b, but the proportion of polygon to polygon, is as ABC to a b c, as now shown: therefore the proportion of polygon to polygon is as the square of AB to the square of ab.

THEO. XXIV.

PL. 1. fig. 8.

Let DHB be a quadrant of a circle described by the radius CB; HB an arc of it, and DH its complement; HL or FC the sine, FH or CL its co-sine, BK its tangent, DI its co-tangent; CK its secant, and CI its co-secant. Fig. 8.

1. The co-sine of an arc is to the sine, as the radius is to the tangent.

2. The radius is to the tangent of an arc, as the co-sine of it is to the sine.

3. The sine of an arc is to its co-sine, as the radius to its co-tangent;

4. Or the radius is to the co-tangent of an arc, as its sine to its co-sine.

5. The co-tangent of an arc is to the radius, as the radius to the tangent.

6. The co-sine of an arc is to the radius, as the radius is to the, secant.

7. The sine of an arc is to the radius, as the tangent is to the secant.

The triangles CLH and CBK, being similar, (by theo. 16.)

1. CL: LH:: CB: BK.

2. Or, CB: BK :: CL: LH.

The triangles CFH and CDI, being similar.

3. CF (or LH): FH :: CD: DI.

4. CD: DI:: CF (or LH): FH.

The triangles CDI and CBK are similar: for the angle CID = KCB, being alternate ones (by part 2. theo. 3.) the lines CB and DI being paralIel: the angle CDI= CBK being both right, and consequently the angle DCI=ČKB, wherefore,

5. DI:CD::CB:BK

And again, making use of the similar triangle CLH and CBK.

6. CL: CB:: CH: CK.

7. HL: CH: BK: CK.

GEOMETRICAL PROBLEMS.

PROB. I.

PL. 2. fig. 7.

To make a triangle of three given right lines BO, LB, LO, of which any two must be greater than the third.

Lay BL from B to L; from B with the line BO, describe an arc, and from L with LO describe another arc; from O, the intersecting point of those arcs, draw BO and OL, and BOL is the triangle required.

This is manifest from the construction.

PROB. II.

PL. 2. fig. 8.

At a point B in a given right line BC, to make an angle equal to a given angle A.

Draw any right line ED to form a triangle, as EAD, take BF=AD, and upon BF make the triangle BFG, whose side BG=AE, and GF=ED (by the last) then also the angle B A; if we suppose one triangle be laid on the other, the sides

=