there will result

Consequently,

AB.AC: DE.DF :: AC2 : DF2.

ABC: DEF:: AC2: DF2.

In like manner it may be shown that

ABC DEF:: AB2: DE2 :: BC2: EF2. Hence,

Any two similar triangles are to each other as the squares of their homologous sides.

Two similar polygons ABCDE, FGHIK, are composed of the same number of triangles which are similar each to each, and similarly situated.

ponding angle F, in the other polygon FGHIK, draw diagonals FH, FI to the other angles.

These polygons being similar, the angles ABC, FGH, which are homologous, must be equal, (Def. 3. 4,) and the sides AB, BC must also be proportional to FG, GH, that is, AB: FG:: BC: GH. Wherefore the triangles ABC, FGH have each an equal angle, contained between proportional sides; hence they are similar; (20. 4;) consequently the angle BCA is equal to GHF. Take these equal angles from the equal angles BCD, GHI; there remains ACD=FHI. But since the triangles ABC, FGH are similar, we have AC : FH :: BC: GH; and since the polygons are similar, BC: GH ::

CD: HI; therefore AC: FH :: CD: HI. But the angle ACD, we already know, is equal to FHI; hence the triangles ACD, FHI have an equal angle in each, included between proportional sides, and are consequently similar. (20. 4.) In the same manner might all the remaining triangles be shown to be similar, whatever be the number of sides in the polygons proposed. Hence,

Similar polygons are composed of the same number of triangles which are similar to each other, and similarly situated.

Scholium. Conversely, if two polygons are composed of the same number of triangles, similar and similarly situated, those two polygons are similar.

For, the similarity of the respective triangles will give the angles ABC=FGH, BCA-GHF, ACD=FHI; hence BCD=GHI, likewise CDE=HIK, &c. Moreover we shall have AB FG :: BC: GH:: AC: FH:: CD: HI, &c., hence the two polygons have their angles equal and their sides proportional; consequently they are similar.

The contours or perimeters of similar polygons are to each other as their homologous sides; and their areas are to each other as the squares of those sides.

are all equal, there

E

K

fore the sum of the antecedents AB+BC+CD, &c.,

the perimeter of the first polygon, is to the sum of the

consequents FG+GH+HI, &c., the perimeter of the second polygon, as any one antecedent is to its consequent; (1.3. Cor.;) therefore, as the side AB is to its corresponding side FG.

Second. Since the triangles ABC, FGH are similar, the triangle ABC FGH :: AC2 : FH2; (25. 4;) and in like manner, from the similar triangles ACD, FHI, we have ACD: FHI: AC2: FH2; therefore, (11. 3,)

ABC FGH :: ACD: FHI.

By the same mode of reasoning, we find

ACD FHI:: ADE: FIK;

:

and so on, if there were more triangles. And from this se ries of equal ratios, we have the sum of the antecedents ABC+ACD+ADE, or the polygon ABCDE, to the sum of the consequents FGH+FHI+FIK, or to the polygon FGHIK, as one antecedent ABC is to its consequent FGH, or as AB2 is to FG2. (1. 3. Cor.) Hence,

The contours or perimeters of similar polygons are to each other as their homologous sides; and their areas are to each other as the squares of those sides.

PROPOSITION XXVIII.

THEOREM.

The segments of two chords AB, CD, which intersect each other in a circle, are reciprocally proportional; that is,

AO DO: CO: OB.

Join AC and BD. In the triangles ACO, BOD the angles at O are equal, being vertical; the angle A is equal to the angle D, because both are inscribed in the same segment; (13. 2. Cor. 1;) A for the same reason the angle C=B; the

B

triangles are therefore similar, and the homologous sides give the proportion,

AO DO CO: OB. Hence,

The segments of any two chords which intersect each other in a circle, are reciprocally proportional.

Cor. Therefore AO.OB DO.CO: hence the rectangle under the two segments of the one chord, is equal to the rectangle under the two segments of the other.

If from the same point without a circle, the secants OB, OC, be drawn terminating in the concave arc BC, the whole secants will be reciprocally proportional to their external segments; that is, OB OC :: OD: OA.

For, joining AC, BD, the triangles OAC OBD have the angle O common; likewise the angle B-C; (13. 2. Cor. 1;) therefore these triangles are similar; and their homologous sides give the proportion, OB OC :: OD: OA. Hence,

If from the same point without a circle, B any two secants be drawn terminating in the

A

concave arc, the whole secants will be reciprocally proportional to their external segments.

Cor. Therefore, the rectangle OA.OB is equal to the rectangle OC.OD.

If from the same point O without a circle, a tangent OA and a secant OC be drawn, the tangent will be a mean proportional between the secant and its external segment; that is, OC: OA : : OA: OD; or, OA2=OC.OD.

For, joining AD and AC, the triangles OAD, OAC have the angle O common ; also the angle OAD, formed by a tangent and a chord, has for its measure half of the arc AD; (14. 2;) and the angle C has the same measure: hence the angle OAD=C; therefore the two triangles are similar; consequently OC: OA: OD; or, (5. 3;) OA-OC.OD.

OA:
Hence,

C

If from the same point without a circle, a tangent and a secant be drawn, the tangent will be a mean proportional between the secant and its external segment.

PROBLEMS

RELATING TO THE FOURTH BOOK.

PROBLEM I.

To divide a given straight line into any number of equal parts, or into parts proportional to given lines.

First. Let it be proposed to divide the line AB into five equal parts. Through the extremity A, draw the indefinite straight line AG; and taking AC of any magnitude, apply it five times upon AG; join the last point of division G and the extremity B, by the straight line GB; then draw CI parallel to GB: AI will be

F

E

D

-K

-L

-M

B