In

the triangles ACD, FHI have an equal angle in each, included between proportional sides, and are consequently similar. the same manner might all the remaining triangles be shewn to be similar, whatever were the number of sides in the polygons proposed: therefore two similar polygons are composed of the same number of triangles similar and similarly situated.

Scholium. The converse of the proposition is equally true: If two polygons are composed of the same number of triangles similar and similarly situated, those two polygons will be si

milar.

For, the similarity of the respective triangles will give the angles ABC-FGH, BCA-ĜHF, 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; hence they are similar.

PROPOSITION XXVII. THEOREM.

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

First. Since, by the nature of similar figures, we have (see the preceding figure) AB: FG :: BC: GH :: CD: HI, &c. we conclude from this series of identical ratios, that 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, therefore as the side AB is to its corresponding side FG.

Secondly. Since the triangles ABC, FGH are similar, we shall have (III. 25.) the triangle ABC: FGH:: AC2 : FH2; and in like manner, from the similar triangles ACD, FHI, we shall have ACD: FHI :: AC2: FH2; therefore by reason of the common ratio, AC2: FH2, we have

ABC: FGH:: ACD: FHI. By the same mode of reasoning, we should find

ACD: FHI::ADE:FIK;

and so on, if there were more triangles. And from this series of identical ratios, we conclude that the sum of the antecedents ABC+ACD+ADE, or the polygon ABCDE, is 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; hence the surfaces of similar polygons are to each other as the squares of the homologous sides.

Cor. If three similar figures were constructed, on the three sides of a right-angled triangle, the figure on the hypotenuse would be equal to the sum of the other two: for the three figures are proportional to the squares of their homologous sides; but the square of the hypotenuse is equal to the sum of the squares of the two other sides; hence, &c.

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 (18. II.); for the same reason the angle C-B; the triangles are therefore similar, and the homologous sides give the proportion AO: DO:: CO: OB.

A

B

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.

PROPOSITION XXIX. THEOREM.

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 (18. II.); these triangles are therefore similar; and their homologous sides give the proportion, OB: OC :: OD: OA.

Cor. The rectangle OA.OB is hence equal to the rectangle OC.OD.

Scholium. This proposition, it may be observed, bears a great analogy to the preceding, and B differs from it only as the two chords AB, CD,

instead of intersecting each other within the circle, cut each other externally. The following proposition may also be regarded as a particular case of the proposition just demonstrated.

PROPOSITION XXX. THEOREM.

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, we shall have OC: OA:: OA: OD; or, which is the same thing, 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 (19. II.) half of the arc AD; and the angle C has the same measure: hence the angle OAD=C; hence the two triangles are similar, and we have the proportion, OC: OA:: OA: OD,

which gives OA2=OC. OD.

PROPOSITION XXXI. THEOREM,*

In a triangle ABC, if the angle A is bisected by the line AD, the rectangle of the sides AB, AC will be equal to the rectangle of the segments BD, DC, together with the square of AD, the bisecting line.

Describe a circle through the three points A, B, C; produce AD till it meets the circumference, and join CE.

The triangle BAD is similar to the triangle EAC; for by hypothesis the angle BAD-EAC; also the angle B-E, since they both have for measure half of the arc AC; hence these triangles are similar, and the homologous sides give the proportion,

B

D

BA: AE:: AD: AC; hence BA.AC-DE.AD; but AE AD+DE, and multiplying each of these equals by AD, we have AE.ADAD+AD.DE; now AD.DE=BD.DC (28. III.); hence finally,

BA.AC=AD2+BD.DC.

This and the three succeeding propositions are not immediately connected with the chain of geometrical investigation. They may be omitted or not, as the reader chooses.-ED.

PROPOSITION XXXII. THEOREM.

In every triangle ABC, the rectangle of the two sides AB, AC is equal to the rectangle contained by the diameter CE of the circumscribed circle, and the perpendicular AD let fall upon the third side BC.

For, joining AE, the triangles ABD, AEC are right-angled, the one at D, the other at E A; also the angle B-E; these triangles are therefore similar, and they give the proportion, AB:CE::AD: AC; and hence AB.AC CE.AD.

Cor. If these equal quantities be multiplied by the same quantity BC there will result AB.AC.BC=CE.AD.BC; now AD.BC is double of the surface of the triangle (6. III.); therefore the product of the three sides of a triangle is equal to its surface multiplied by twice the diameter of the circumscribed

circle.

The product of three lines is sometimes called a solid, for a reason that shall be seen afterwards. Its value is easily conceived, by imagining that the lines are reduced into numbers, and multiplying these numbers together.

Scholium. It may also be demonstrated, that the surface of a triangle is equal to its perimeter multiplied by half the radius of the inscribed circle.

For the triangles AOB, BOC, AOC, which have a common vertex at O, have for their common altitude the radius of the inscribed circle; hence the sum of these triangles will be equal to the sum of the bases AB, BC, AC, multiplied by half the radius OD;

A

hence the surface of the triangle ABC is equal to the perimeter multiplied by half the radius of the inscribed circle.

PROPOSITION XXXIII. THEOREM.

In every inscribed quadrilateral ABCD, the rectangle of the two diagonals AC, BD is equal to the sum of the rectangles of the opposite sides; so that we have

AC.BD=AB.CD+AD.BC.

Take the arc CO=AD, and draw BO meeting the diagonal AC in I.

K

The angle ABD CBI, since the one has for its measure half of the arc AD, and the other half of CO equal to AD; the angle ADB BCI, because they are A both inscribed in the same segment AOB; hence the triangle ABD is similar to the triangle IBC, and we have the proportion AD: CI:: BD: BC; hence AD.BC=CI.BD. Again, the triangle ABI is similar to the triangle BDC; for the arc AD being equal to CO, if OD be added to each of them, we shall have the arc AO=DC; hence the angle ABI is equal to DBC; also the angle BAI to BDC, because they are inscribed in the same segment; hence the triangles ABI, DBC are similar, and the homologous sides give the proportion, AB: BD :: AI: CD; hence AB.CD-AI.BD.

Adding the two results obtained, and observing that AI.BD +CI.BD=(AI+CI).BD=AC.BD, we shall have AD.BC+

AB.CD=AC.BD.

Scholium. Another theorem concerning the inscribed quadrilateral be demonstrated in the same manner:

may

The similarity of the triangles ABD and BIC gives the proportion BD: BC:: AB: BI; hence BI.BD=BC.AB. If CO be joined, the triangle ICO, similar to ABI, will be similar to BDC, and will give the proportion BD: CO: : DC : OI; hence OI.BD=CO.DC, or, because COAD,OI.BD-AD.DC. Adding the two results, and observing that BI.BD+OI.BD is the same as BO.BD, we shall have BO.BD=AB.BC+AD.DC. If BP had been taken equal to AD, and CKP been drawn, a similar train of reasoning would have given us

CP.CA=AB.AD+BC.CD.

But the arc BP being equal to CO, if BC be added to each of them, it will follow that CBP-BCO; the chord CP is therefore equal to the chord BO, and consequently BO.BD and CP.CA are to each other as BD is to CA; hence,

BD: CA: AB.BC+AD.DC: AD.AB+BC.CD.