THEOREM XXV.

C

The area of a triangle is equal to the product of the three sides divided by twice the diameter of its circumscribed circle. Let K be the centre of the circumscribed circle. Draw the diameter CKD, and the chord AD, and the perpendicular CH, which will be the altitude of the triangle.

A

D

K

B

H

The two triangles CAD, CHB are rightangled, the one at A (B. II., T. X., Cor. I.), and the other at H; moreover, the angles at D and B are equal, being inscribed in the same segment ADBC. Hence these triangles are similar, and their homologous sides being proportional, we have

' in which, substituting the value of CH just found, we have

ABC:

=

ABX CBX AC
2CD

RATIOS BETWEEN THE AREAS OF SIMILAR FIGURES

THEOREM XXVI.

The areas of two triangles which have an equal angle, are proportional to the rectangles of the sides containing the equal angle.

A A

ABC: ADE:: ABX AC: AD × AE.

For, joining D and C, we have, since triangles of the same altitude are to each other as their bases,

ABC: ADC::AB: AD,

ADC: ADE:: AC: AE.

Multiplying together the corresponding terms of these proportions, and omitting the common term ADC which enters into the antecedent and consequent of the first couplet, we have

ABC: ADE:: AB × AC: AD × AE.

Cor. Whenever the side DE of the second triangle is parallel to BC of the first, the triangle ADC is a mean proportional between the triangles ABC and ADE.

In effect we have, in this case (T. III.),

AB: AD:: AC: AE;

so that the two first proportions of the preceding Theorem have equal ratios, and give

ABC:ADC::ADC:ADE.

THEOREM XXVII.

The areas of similar triangles are proportional to the squares

of their homologous sides.

These triangles being similar, are

equiangular (T. V.).

Hence, by the preceding Theorem,

we have

B

ABC: A'B'C':: AB × AC: A'B' x A'C';

but since the triangles are similar, we also have

AC: AC':: AB: A'B'.

Multiplying the corresponding terms of these two proportions, omitting the common factor AC which enters into the antecedents, and the common factor A'C' which enters into the consequents, we have

ABC: A'B'C':: AB2: A'B'2.

THEOREM XXVIII.

The perimeters of two similar polygons are proportional to their homologous sides. And their areas are proportional to the squares of these sides.

First. By reason of the similarity of these polygons, we have (T. IX.),

AB: A'B':: BC: B'C':: CD: C'D'::, etc.

Now the sum of these antecedents, AB+BC+CD+, etc., which makes the perimeter of the first polygon, is to the sum of their consequents A'B' + B'C' + C'D' +, etc., which makes the perimeter

E

D

B

E'

D'

C

F

B'

of the second polygon, as any one antecedent is to its corresponding consequent, and therefore as AB is to A'B'.

Secondly. Since these polygons are similar, they are each composed of the same number of similar triangles, which, compared together, are each to each, in the same ratio, that of the squares of the homologous sides. It follows, then, that the sum of the triangles which compose the first polygon, is to the sum of the triangles which compose the second polygon, as the squares of the homologous sides of the two polygons. That is, we shall have

ABCDEF: A'B'C'D'E'F' :: AB2: A'B'.

Cor. The perimeters of similar figures are proportional to their homologous lines, and their areas are proportional to the squares of those lines.

For, by the definition of similar figures, the ratio of any two homologous lines, which is called the ratio of similitude, is constantly the same.

COMPARISON OF SQUARES CONSTRUCTED ON CERTAIN

LINES.

THEOREM XXIX.

The square constructed on the hypotenuse of a right-angled triangle, is equivalent to the sum of the squares constructed respectively on the other two sides.

This Theorem is not a fundamental one, like Theorem XIV., but its importance and its fecundity have given rise to several demonstrations founded solely upon the equality and equiva lence of figures. We shall confine ourselves to the one generally given in works of geometry.

The three squares being constructed without the triangle ABC, we draw from the right angle at A the line AL perpendicular to the hypotenuse, and produce it to meet DE at M, thus dividing the square BCED into the two rectangles BDML, LCEM. We also draw AD and CF.

Now, the angles ABD, FBC are equal, being each composed

F

G

A

B

L

D

M

K

E

same square, so also are Hence the two triangles But the triangle ABD have the same base BD

of a right angle and of the common angle ABC; the lines AB and BF are equal, being sides of the BC and BD equal, for a like reason. BAD, BFC are equal (B. I., T. XX.). is half the rectangle BLMD, since they and the same altitude BL (T. XIX.). For the same reason the triangle BFC is half the square ABFG, for they have the same base BF and the same altitude AB; consequently,

BLMD = ABFG.

We can prove in the same manner, by drawing AE and BI,

that

LMEC = ACIK;

and as the square BDEC is equivalent to BLMD+LMEC there will result

M

square BDEC = square ABFG+ square ACIK. Cor. I. The square MNPQ, constructed on the diagonal BD or AC of a square ABCD, is double the square itself.

B

D

P

This is obvious from the simple inspection of the figure. In effect, the four squares AKBM, AKDN, BKCQ, DKCP, are equal, and respectively the doubles of the triangles AKB, AKD, BKC, DKC, of which the sum is equal to the square ABCD.

Cor. II. We have already shown that the squares ABFG, ACIK (see first figure) are respectively equivalent to the rectangles BLMD, LMEC. And since these rectangles and the square BCED have the common altitude BD, they are to each other as their bases BL, LC, and BC (T. XX.). Hence we obtain this relation:

AB2: AC2: BC2::BL:LC : BC.

Cor. III. Since the two squares ABFG, ACIK are respectively equivalent to the rectangles BLMD, LMEC, we have AB2 = BC × BL, AC2 = BC × LC;

from which we deduce these two proportions:

BC: AB:: AB: BL, BC: AC:: AC: LC.

These two proportions, together with the relations of the preceding Corollary, correspond in all respects with the third portion of Theorem XIII., and Scholium I. of Theorem XIV. We have thus, by another method, established properties already demonstrated.

Cor. IV. In short, if on the three sides of a right-angled triangle ABC, we suppose three similar polygons to be constructed, as these polygons will be proportional to the squares of their homologous sides (T. XXVIII.), and we have this relation between these sides:

BC2AB+AC2;

it follows, that, of the three similar polygons, the polygon constructed on the hypotenuse is equivalent to the sum of the polygons constructed on the sides containing the right angle.