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284. If the polygons ABCDE and a b c d e, (fig. 154,) are similar to each other, angle A= angle a, angle B=b, &c., &c.; the two polygons will be divided by homologous diagonals into similar triangles; that is, ABC a b c; ADC o a d c, and ADE ∞ a de.

Dem. Angle ABC a b c and the sides AB and BC are proportional to the homologous sides a b and b c, therefore tringle ABC co a b c.

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Again, triangle ABC a b c, therefore angle ACB= a cb; angle BCD= b c d ; therefore angle BCD — ACB ACD = bcd-acbac d. Since triangle ABC

cs abc,

therefore BC: bc AC: ac

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In a like manner a ed.

consequently triangle ACD ∞ a c d.
it may be demonstrated that triangle AED

285. (Fig. 154.) If the triangle ABC ∞ a b c, ACD ∞ a c d and AED ca e d, and these triangles are placed together in a similar order, so that the sides and angles of one set shall correspond in position with the equal angles and proportional sides of the other set, then the entire figure ABCDE will be similar to the entire figure a b c d e.

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Dem. Triangle ABC co a b c; ACD co a cd; ADE ade; consequently angle ABC-a bc; angle BCA bca; angle ACD-a cd; therefore angle BCA+

ACD=BCD=angle bca+acd=bcd. The equality of the other homologous angles in the two polygons can be demonstrated in a similar manner.

Since the triangles are similar, we have the following proportions among the sides;

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Consequently in the two polygons not only are the homologous angles equal, but the homologous sides are proportional, therefore the polygons are similar.

286. (Fig. 155.) To construct upon the line FG a figure which shall be similar to a given figure ABCDE.

Sol. 1. Divide ABCDE by diagonals into 3 triangles; then beginning upon the line FG, construct 3 triangles similar to those of the polygon, and similarly disposed. For example, suppose FG to be homologous to AB, and upon it construct a triangle similar to ABC; then upon the side of this triangle which is homologous to side AC, construct a triangle similar to ADC; then upon the side of this second triangle which is homologous to side AD, construct a triangle similar to ADE,

and this will complete the figure, which will be similar to ABCDE.

Sol. 2. Produce AB to H, taking AH = FG. Produce the diagonals AC and AD, and the side AE. Draw HI parallel to BC, IK parallel to CD, KL parallel to DE. The required figure will be constructed. If FG is less than AB, so that the point H falls between the points A and B, the problem would be solved by drawing lines parallel to the sides of the given figure.

287. To construct a diagram of a field upon a reduced scale.

If the field can be measured in all directions from the inside, then divide it by diagonals into rectilineal triangles and measure all the sides of these triangles. Represent these triangles upon paper on a reduced scale in the same relative position that they occupy in the field, and thus the problem will be solved.

If the field cannot be measured from the inside, (as a piece of woodland,) but can be readily reached upon the outside, then measure all the sides of the perimeter, and all the angles. Draw the sides on a reduced scale upon paper, joining them together in their natural order, so as to form together the measured angles.

288. Suppose ABCDE ∞ a b c d e, what is the ratio of AB+BC+CD+DE+ EA to a b+b c+cd+ de+ea? (fig. 154.)

Answer.

AB: a b BC: b c = CD: cd=DE: de=EA: e a. This is called a continued proportion, being a series of equal ratios.

In every continued proportion the sum of all the ante

cedents is to the sum of all the consequents as one antecedent is to its consequent.

Therefore

AB+BC+CD+DE+EA: ab+b c + cd+de+ea=AB: a b = BC: bc; that is, the perimeters of similar figures are to each other as their homologous sides.

289. What ratio does the surface of ABCDE bear to that of a bede? (Fig. 154.)

Sol. Divide the similar figures ABCDE and abcde by homologous diagonals into similar triangles. We then have.

2 -2

triangle ABC: triangle a b c = AC:ac (282)

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triangle ACD: triangle a c d― AC: ac

therefore,

triangle ABC: triangle abc triangle ACD: triangle a cd. Again,

triangle ACD: triangle acd=AD: ad

2

triangle ADE: triangle a de— AD3:
= AD2: ad2

therefore,

triangle ACD: triangle acd=triangle ADE: triangle ade. Because

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triangle ABC: triangle abc triangle ACD: triangle a cd triangle ADE: triangle a de

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therefore,

triangle ABC+ triangle ACD+ triangle ADE: triangle abc+triangle acd+triangle a de= triangle ABC: triangle abc triangle ADC: triangle a dc triangle

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2

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ADE : triangle a de=AB: ab=BC: be, &c., that is, The surfaces of similar figures are to each other as the squares of their homologous sides.

290. Upon the 3 sides of a right-angled triangle construct similar figures, in which the sides of the triangle shall be similarly disposed, then the figure constructed upon the hypothenuse will be equal to the sum of the figures constructed upon the two sides which include the right angle. Figure C=fig. A+ fig. B, (fig. 156.)

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Dem. Figure A: fig. B-side a2: side 2,

therefore

Again,

therefore

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a

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-2

B:C=b : C

-2-2

A+B:C= a + b : c ;

2+b2=c2, therefore A+B=C.

291. To construct a figure which shall be similar to two given similar figures ABCDE and abcde, and shall be equivalent to both taken together, (fig. 157.)

Sol. Make a right angle. Take one of the sides f g equal to a side AB of one of the given figures, and the other side ƒh equal to the homologous side, a b, of the other given figure. Draw the hypothenuse h g, and upon this line h g construct a figure similar to ABCDE and a bede, and in which the side h g shall be homologous to the sides AB and ab. It has been demonstrated in the preceding proposition that the figure constructed upon h g will be equivalent to both the figures constructed upon the sides fh and fg, or their equals AB and a b, taken together.

5. CIRCLES.

292. If two chords cut each other in a circle, the parts are reciprocally proportional, (fig. 158.)

The part AE of the chord AB is to the part CE of the chord CD, not as the second part EB of the first

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