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Cor. Hence the two triangles would be equivalent, if the rectangle AB.AC were equal to the rectangle AD.AE, or if we had AB AD :: AE : AC; which would happen if DC were paralel to BE.


Two similar triangles are to each other as the on their homologous sides.

Let ABC, DEF, be two similar triangles, having the angle A equal to D, and the angle B-E.

Then, first, by reason of the equal an- G gles A and D, according to the last proposition, we shall have

Also, because the triangles are similar,
AB: DE :: AC: DF,


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And multiplying the terms of this proportion by the corresponding terms of the identical proportion,



there will result





AC: DE2.

Therefore, two similar triangles ABC, DEF, are to each other as the squares described on their homologous sides AC, DF, or as the squares of any other two homologous sides.


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

Let ABCDE, FGHIK, be two similar polygons. From any angle A, in the polygon ABCDE, B draw diagonals AC, AD to the other angles. From the homologous 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, and the sides AB, BC, must also be proportional to FG, GH, that is, AB : FG :: BC: GH (Def. 1.). Wherefore the triangles ABC, FGH, have each an equal angle, contained between proportional sides; hence they are similar (Prop. XX.); therefore the angle BCA is equal to GHF. Take away these equal angles from the equal angles BCD, GHI, and 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; hence 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 (Prop. XX.). In the same manner it might be shown that all the remaining triangles are similar, whatever be 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 similar. For, the similarity of the respective triangles will give the angles, ABC=FGH, BCA=GHF, ACD=FÍI: 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 the homologous sides: and the areas are to each other as
squares described on those sides.


First. Since, by the nature of similar figures, B we have AB : FG :: BC GH:: CD: HI, &c. we conclude from this series of equal ratios that the sum of the antecedents AB+BC+CD,

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&c., which makes up the perimeter of the first polygon, is to the sum of the consequents FG+GH+HI, &c., which makes up the perimeter of the second polygon, as any one antecedent is to its consequent; and therefore, as the side AB is to its corresponding side FG (Book II. Prop. X.).


Secondly. Since the triangles ABC, FGH are similar, we shall have the triangle ABC FGH AC: FH2 (Prop. XXV.); 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



By the same mode of reasoning, we should find

and so on, if there were more triangles. And from this series of equal 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 (Prop. XXV.); hence the areas of similar polygons are to each other as the squares described on the homologous sides.

If three similar figures were constructed, on the three sides of a right angled triangle, the figure on the hypothenuse would be equivalent 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 hypothenuse is equivalent to the sum of the squares of the two other sides; hence, &c.

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

Let the chords AB and CD intersect at O: then will AO DO: OC : OB. Draw 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 (Book III. Prop. XVIII. Cor. 1.); for the same reason the angle CB; the triangles are therefore similar, and the homologous sides give the proportion AO: DO: CO: OB.

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, two secunts be drawn terminating in the concave arc, the whole secants will be reciprocally proportional to their external segments.

Let the secants OB, OC, be drawn from the point 0: then will

OB OC :: OD : OA.


For, drawing AC, BD, the triangles OAC, OBD have the angle O common; likewise the angle B=C (Book III. Prop. XVIII. Cor. 1.); these triangles are therefore similar; and their homologous sides give the proportion,

OB: OC :: OD : OA.

Cor. Hence the rectangle OA.OB is equal to the rectangle OC.OD.


Scholium. This proposition, it may be observed, bears a great analogy to the preceding, and differs from it only as the two chords AB, CD, instead of intersecting each other within, cut each other without the circle. The following proposition may also be regarded as a particular case of the proposition just demonstrated.


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.

From the point O, let the tangent OA, and the secant OC be be drawn ; then will


OA: OD, or OA2-OC.OD. For, drawing AD and AC, the triangles O 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 (Book III. Prop. XXI.); and the angle Chas the same measure: hence the angle OAD= A C; therefore the two triangles are similar, and we have the proportion OC: OA :: AO OD, which gives OA2=OC.OD.


If either angle of a triangle be bisected by a line terminating in the opposite side, the rectangle of the sides including the bisected angle, is equivalent to the square of the bisecting line together with the rectangle contained by the segments of the third side.

In the triangle BAC, let AD bisect the angle A; then will AB.AC=AD2+BD.DC.

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




The triangle BAD is similar to the triangle EAC; for, by hypothesis, the angle BAD EAC; also the angle B=E, since they are both measured by half of the arc AC; hence these triangles are similar, and the homologous sides give the proportion BA : AE :: AD : AC; hence BA.AC=AE.AD; but AE=AD+DE, and multiplying each of these equals by AD, we have AE.AD=AD2+ AD.DE; now AD.DE=BD.DC (Prop. XXVIII.); hence, finally,


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