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Plate II.

Whence also the parallelograms ABCF and BDEC, being (by cor. 2. theo. 12.) the doubles of the triangles, are likewise as their bases. Q. E. D.

Note. Wherever there are several quantities connected with the sign the conclusion is always drawn from the first two and last two proportionals.


Triangles ABC, DEF, standing upon equal bases AB and DE, are to each other as their altitudes CG and FH. fig. 2.

Let BI be perpendicular to AB and equal to CG, in which let KB-FH, and let AI and AK be drawn.

The triangle AIB=ACB (by cor. to theo. 13.) and AKB-DEF; but (by theo. 18.) BI: BK:: ABI: ABK. That is, CG: FH:: ABC: DEF,

Q. E. D.


If a right line BE be drawn parallel to one side of a triangle ACD, it will cut the two other sides proportionally, viz. AB: BC:: AE: ED. fig. 3.

Draw CE and BD; the triangles BEC and EBD being on the same base BE and under the same parallel CD, will be equal (by cor. to theo. 13.)_ therefore (by theo. 18.) AB: BC:: (BEA: BEC or BEA: BED) :: AÉ: ED. Q. E. D.

Plate II.

Cor. I. 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.


If two triangles ABC, ADE, have angle BAC in one, equal to one 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. Fig. 4.

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

Because (by the first cor. to the foregoing theo.) Ab: Ad :: (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 triangles Adc ADE, therefore the angle Ade = D and Aed E, but since ed and BC are parallel (by part 3. theo. 3.) Ade B, and Aed therefore B-D and C=E. Q. E. D.




Plate II.


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. Fig. 5.

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 DEF, by cor. 2. theo. 12. Q. E. D.


Like polygons ABCDE, a b c d e, are in a duplicate proportion to that of 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. Fig. 6.

Draw AD, AC, ad, ac.

Plate II.

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-e da. If therefore from the equal angles A, and a, we take the equal ones EAD + BAČ=ead, + bac the remaining angle DAC-dac, and if from the equal angles D and d, EDA-ed a be taken, we shall have ADC—a dc: and in like manner if from C and c be taken BCA-b ca, 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: adc :: the square of AC, to the square of ac; therefore from equality of proportions ABC: a b c :: ADC: ad c, 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: 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 abc, 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.

Plate II.


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 ils secant, and CI its co-secant. Fig. 8.

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

2. Radius is to the tangent of an arc, as the cosine of it is to the sine.

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

4. Or 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 radius, as radius to the tangent.

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

7. The sine of an arc is to 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.

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