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similar. In the same manner it may be shown that the triangles DAC and BAC are similar; hence all the triangles are equiangular and similar.

Secondly. The triangles BAD, BAC, being similar, their homologous sides are proportional. But BD in the small triangle, and BA in the large one, are homologous sides, because they lie opposite the equal angles BAD, BCA; the hypothenuse BA of the small triangle is homologous with the hypothenuse BC of the large triangle: hence the proportion BD : BA: BA: BC. By the same reasoning, we should find DC AC AC: BC; hence, each of the sides AB, AC, is a mean proportional between the hypothenuse and the segment adjacent to that side.

Thirdly. Since the triangles ABD, ADC, are similar, by comparing their homologous sides, we have BD: AD: : AĎ : DC; hence, the perpendicular AD is a mean proportional between the segments BD, DC, of the hypothenuse.

Scholium. Since BD: AB :: AB BC, the product of the extremes will be equal to that of the means, or AB2=BD.BC. For the same reason we have AC2=DC.BC; therefore AB2+ AC2=BD.BC+DC.BC= (BD+DC).BC=BC.BC=BC2; or the square described on the hypothenuse BC is equivalent to the squares described on the two sides AB, AC. Thus we again arrive at the property of the square of the hypothenuse, by a path very different from that which formerly conducted us to it and thus it appears that, strictly speaking, the property of the square of the hypothenuse, is a consequence of the more general property, that the sides of equiangular triangles are proportional. Thus the fundamental propositions of geometry are reduced, as it were, to this single one, that equiangular triangles have their homologous sides proportional.

It happens frequently, as in this instance, that by deducing consequences from one or more propositions, we are led back to some proposition already proved. In fact, the chief charac teristic of geometrical theorems, and one indubitable proof of their certainty is, that, however we combine them together, provided only our reasoning be correct, the results we obtain are always perfectly accurate. The case would be different, if any proposition were false or only approximately true: it would frequently happen that on combining the propositions together, the error would increase and become perceptible. Examples of this are to be seen in all the demonstrations, in which the reductio ad absurdum is employed. In such demonstrations, where the object is to show that two quantities are equal, we proceed by showing that if there existed the smallest

inequality between the quantities, a train of accurate reasoning would lead us to a manifest and palpable absurdity; from which we are forced to conclude that the two quantities are equal.

Cor. If from a point A, in the circumference

of a circle, two chords AB, AC, be drawn to the extremities of a diameter BC, the triangle BAC will be right angled at A (Book III. Prop. B D XVIII. Cor. 2.); hence, first, the perpendicular AD is a mean proportional between the two segments BD, DC, of the diameter, or what is the same, AD-BD.DC.

Hence also, in the second place, the chord AB is a mean proportional between the diameter BC and the adjacent segment BD, or, what is the same, AB2= BD.BC. In like manner, we have AC2-CD.BC; hence AB2: AC2 :: BD : DC; and comparing AB and AC2, to BC2, we have AB2: BC2:: BD : BC, and AC BC2: DC: BC. Those proportions between the squares of the sides compared with each other, or with the square of the hypothenuse, have already been given in the third and fourth corollaries of Prop. XI.


Two triangles having an angle in each equal, are to each other as the rectangles of the sides which contain the equal angles.

In the two triangles ABC, ADE, let the angle A be equal to the angle A; then will the triangle


Draw BE. The triangles ABE, ADE, having the common vertex E, have the same altitude, and consequently are to each other as their bases (Prop. VI. Cor.): that is,

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Multiply together the corresponding terms of these proportions, omitting the common term ABE; we have


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 parallel to BE.


Two similar triangles are to each other as the squares described 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,





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

there will result



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

ABC: DEF :: AC2: DF2.

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 an-






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 FHF; 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=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 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,





&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 AC2: 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 poly. gons are to each other as the squares described on the homologous sides.

Cor. 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 in a circle, are reciprocally proportional.

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